## Selina Concise Physics Class 10 ICSE Solutions Refraction of Light at Plane Surfaces

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 4 Refraction of Light at Plane Surfaces. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 4 Refraction of Light at Plane Surfaces

Exercise 4(A)

Solution 1.

The change in the direction of the path of light, when it passes from one transparent medium to another transparent medium, is called refraction of light.

Solution 2.

Solution 3.

The ray of light which is incident normally on a plane glass slab passes undeviated. That is such a ray suffers no bending at the surface because here the angle of incidence is 0°. Thus if angle of incidence ∠i = 0°, then the angle of refraction ∠r = 0°. And the angle of deviation of the ray will also be 0°.

Solution 5.

The refraction of light (or change in the direction of path of light in other medium) occurs because light travels with different speeds in different media. When a ray of light passes from one medium to another, its direction (except for ∠i = 0°) changes because of change in its speed.

Solution 6.

Air is a rarer medium while water is denser than air with refractive index of 1.33. Therefore when light ray will travel from air to water it will bend towards the normal.

Solution 7.

Speed, intensity and wavelength

Solution 8.

The Snell’s laws of refraction are:

1. The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
2. The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for the pair of the given media.

Solution 9.

The refractive index of second medium with respect to first medium is defined as the ratio of the sine of angle of incidence in the first medium to the sine of the angle of refraction in the second medium.
Refractive index of a medium is always greater than 1 (it cannot be less than 1) because the speed of light in any medium is always less than that in vacuum.

Solution 10.

Denser medium has a higher refractive index and therefore the speed of light in such medium is lower in comparison to the speed of light in a medium which has a lower refractive index.

Solution 11.

The speed of light decreases when it enters from a rarer medium to denser medium and increases when it enters from a denser medium to rarer medium.
Therefore, the speed of light increases when light ray passes from water to air and the speed of light decreases when light ray passes from water to glass.

Solution 12.

(a) Air (its refractive index is less than that of water)
(b) Glass (its refractive index is more than that of water)

Solution 13.

The refractive index of glass is 1.5 for white light means white light travels in air 1.5 times faster than in glass.

Solution 14.

Solution 16.

(a) The least for red colour and (b) the most for violet colour.

Solution 17.

1. Nature of a medium i.e. its optical density (e.g. μ= 1.5, μ= 1.33) – Smaller the speed of light in a medium relative to air, higher is the refractive index of that medium.
2. Physical condition such as temperature – with increase in temperature, the speed of light in medium increases, so the refractive index of medium decreases.

Solution 18.

Refractive index of a medium decreases with increase in wavelength of light.
Refractive index of a medium for violet light (least wavelength) is greater than that for red light (greatest wavelength).

Solution 19.

Refractive index of a medium decreases with the increase in temperature.
With increase in temperature, the speed of light in that medium increases; thus, the refractive index (= velocity of light in vacuum/velocity of light in medium) decreases.

Solution 21.

Solution 22.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

(i) The glass piece is not seen when the refractive index of liquid becomes equal to the refractive index of glass.
(ii) Light of a single colour is used because the refractive index of a medium (glass or liquid) is different for the light of different colours.

Solution 28.

When a ray of light from lighted candle fall on the surface of a thick plane glass mirror, a small part of light (nearly 4%) is reflected forming first image which is faint virtual image, while a large part of light (nearly 96%) is refracted inside the glass. This ray is now strongly reflected back by the silvered surface inside the glass. This ray is then partially refracted in air and this refracted ray forms another virtual image. This image is the brightest image because it is due to the light suffering a strong reflection at the silver surface.

Solution 29.

(a) When light travels from a rarer to a denser medium, its speed decreases
(b) When light travels from a denser to a rarer medium, its speed increases
(c) The refractive index of glass with respect to air is 3/2. The refractive index of air with respect to glass will be 2/3

Solution 1 (MCQ).

The ray of light bends towards the normal.
Reason: As the speed of light decreases in the denser medium, it bends towards the normal.

Solution 2 (MCQ).

(a) 0°
Reason: A ray of light which is incident normally (i.e. at angle of incidence = 0°) on the surface separating the two media, passes undeviated.

Solution 3 (MCQ).

Diamond
Reason: As the speed of light in diamond is the least, diamond has the highest refractive index.

Numericals

Solution 1.

Solution 2.

Solution 3.

Solution 4.

Exercise 4(B)

Solution 1.

Solution 2.

Solution 3.

The angle between the direction of incident ray and the emergent ray, is called the angle of deviation.

Solution 4.

Angle of deviation is the angle which the emergent ray makes with the direction of incident ray.

Solution 5.

In a prism the ray of light suffers refraction at two faces. The prism produces a deviation at the first surface and another deviation at the second surface. Thus a prism produces a deviation in the path of light.
The value of the angle of deviation (or the deviation produced by a prism) depends on the following four factors:
(a) the angle of incidence (i),
(b) the material of prism(i.e., on refractive index μ.),
(c) the angle of prism (A),
(d) The colour or wavelength (λ) of light used.

Solution 6.

As the angle of incidence increases, the angle of deviation decreases first and reaches to a minimum value (δm) for a certain angle of incidence. By further increasing the angle of incidence, the angle of deviation is found to increase.
Variation of angle of deviation (δ) with angle of incidence (i):

Solution 7.

False.
With the increase in the angle of incidence, the deviation produced by a prism first decreases and then increases.
A given prism deviates the violet light most and the red light least.

Solution 8.

Solution 9.

Changes in the angle of deviation as we increase

1. The wavelength of incident light
As we increase the wavelength, angle of deviation decreases.
2. The refracting angle of the prism
The angle of deviation increases with the increase in the angle of prism.

Solution 10.

The relation between the angle of incident (i), angle of emergence (e), angle of prism (A) and angle of deviation (δ) for a ray of light passing through an equilateral prism is δ = (i + e) – A

Solution 11.

(i) i2 = i1
(ii) Angle of deviation is minimum
Explanation: In minimum deviation position, the refracted ray inside the prism travels parallel to it if the prism is equilateral and the angle of incidence is equal to the angle of emergence.

Solution 12.

In case of an equilateral prism, when the prism is in the position of minimum deviation δ = δmin, the angle of incidence i1 is equal to the angle of emergence i2.

Solution 13.

Solution 14.

(i) Violet colour will deviate the most and (ii) Red colour will deviate the least.

Solution 15.

B made of flint glass. Because it has higher refractive index.

Solution 16.

The angle of deviation (δ) increases with the increase in the angle of prism (A).

Solution 17.

Let two rays OA and OL from a source O are incident on the prism. They are refracted along AB and LM from first face of the prism. These two rays again refract from the second face of the prism emerge out along BC and MN respectively such that they appear to come from I.

Solution 18.

(a) If the incident ray normal to prism then angle of incidence is 0o.
(b) In this case the angle of refraction from the first face r1= 0o.
(c) As the prism is equilateral so A=60o and r1=0o. So at the second face of the prism, the angle of incidence will be 60o.
(d) No the light will not suffer minimum deviation.

Solution 19.

Solution 1 (MCQ).

The light ray bends at both the surfaces of prism towards its base.

Solution 2 (MCQ).

The deviation produced by the prism does not depend on the size of prism.

Numericals

Solution 1.

Solution 2.

Exercise 4(C)

Solution 1.

Solution 2.

Consider a ray of light incident normally along OA. It passes straight along OAA’. Consider another ray from O (the object) incident at an angle i along OB. This ray gets refracted and passes along BC. On producing this ray BC backwards, it appears to come from the point I, and hence, AI represents the apparent depth, which is less than the real depth AO.

Solution 3.

The depth of the tank appears to be lesser than its real depth. This happens due to the refraction of light from a denser medium (water) to a rarer medium.

Solution 4.

Let any object B is at the bottom of a pond. Consider a light ray BC from the object that moves from water to air. After refraction from the water surface, the ray moves away from the normal N along the path CD. The produce of CD appears from the point B’ and a virtual image of the object at B appears at B’.

Solution 5.

A stick partially immersed in water in a glass container appears bent or raised as shown in figure above. This happens because the rays appear to come from P’ (which is the virtual image of the tip P of the stick) due to refraction from denser medium (water) to rarer medium (air) at the surface separating two media.

Solution 6.

Let the fish is looking from the point O. As the ray OP emerges out from water to air, it will bend away from the normal MN because air is a rarer medium in comparison of water. But if we extend ray OP then it will meet at Q due to which the plant AB will look taller than its actual height.

Solution 7.

(i) Part of the pencil which is immersed in water will look short and raised up.
(ii) The phenomena which is responsible for the above observation is refraction of light.
(iii) The required figure is

Solution 8.

The factors on which the magnitude of shift depends are:

1. The refractive index of the medium,
2. The thickness of the denser medium and
3. The colour (or wavelength) of incident light.

The shift increase with the increase in the refractive index of medium. It also increases with the increase in thickness of denser medium but the shift decreases with the increases in the wavelength of light used.

Solution 1 (MCQ).

Refraction of light
Hint: When a ray of light travels from denser to rarer medium, it moves away from the normal.

Solution 2 (MCQ).

The shift is maximum for violet light.
Hint: The shift is maximum for violet light because the refractive index of glass is the most for the violet light and apparent = (real depth)/(refractive index)

Numericals

Solution 1.

Solution 2.

Solution 3.

Exercise 4(D)

Solution 1.

Critical angle: The angle of incidence in the denser medium corresponding to which the angle of refraction in the rarer medium is 90° is called the critical angle.

Solution 2.

Solution 3.

Solution 4.

The critical angle for diamond is 24°. This implies that at an incident angle of 24° within the diamond the angle of refraction in the air will be 90°. And if incident angle will be more than this angle then the ray will suffer Total internal reflection without any refraction.

Solution 5.

When a ray is incident from a denser medium to a rarer medium at angle equal to critical angle (i = ic), the angle of refraction becomes 90°.

Solution 6.

The factors which affect the critical angle are:

1. The colour (or wavelength) of light, and
2. The temperature

Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus critical angle increases with increase in wavelength of light.
Effect of temperature: The critical angle increases with increase in temperature because on increasing the temperature of medium, its refractive index decreases.

Solution 7.

As the wavelength decreases (or increases) refractive index becomes more (or less) and critical angle becomes less (or more).

1. For red light the critical angle will be more than 45° and
2. For blue light the critical angle will be less than 45°.

Solution 8.

(a) Total internal reflection: It is the phenomenon when a ray of light travelling in a denser medium, is incident at the surface of a rarer medium such that the angle of incidence is greater than the critical angle for the pair of media, the ray is totally reflected back into the denser medium.
(b) The two necessary conditions for total internal reflection are:

1. The light must travel from a denser medium to a rarer medium.
2. The angle of incidence must be greater than the critical angle for the pair of media.

(c) When incidence angle is more than critical angle i.e., in case of total internal reflection.

Solution 9.

(a) Total internal reflection occurs when a ray of light passes from a denser medium to a rarer medium.
(b) Critical angle is the angle of incidence in denser medium for which the angle of refraction in rarer medium is 90°.

Solution 10.

True

Solution 11.

(b) If refractive angle, r = 90°, the corresponding angle of incidence, i will be equal to critical angle.
(c) If the angle of incidenceexceeds the value of i obtained in part (b) (i.e., critical angle), total internal reflection will occur.

Solution 12.

Solution 13.

(a) Critical angle
Hint: The angle of incidence in the denser medium for which the angle of refraction in rarer medium is 90° is called the critical angle.

(b) 90°

(c) Total internal reflection.
Hint: When the angle of incidence is greater than the critical angle, the phenomenon of total internal reflection occurs due to which the ray of light is not refracted but is reflected back in the same medium.

(d) For the ray PR, the angle of incidence is less than the critical angle (i.e. ∠PQS ); hence, at the interface of two media as per the laws of reflection, ray PR suffers partial reflection and refraction.

Solution 14.

Solution 15.

Solution 16.

A prism having an angle of 90° between its two refracting surfaces and the other two angles each equal to 45°, is called a total reflecting prism. The light incident normally on any of its faces, suffers total internal reflection inside the prism.
Due to this behavior, a total reflecting prism is used to produce following three actions:

1. To deviate a ray of light through 90°,
2. To deviate a ray of light through 180°, and
3. To erect the inverted image without producing deviation in its path.

Solution 17.

As shown in diagram, a beam of light is incident on face AB of the prism normally so it passes undeviated and strikes the face AC where it makes an angle of 45° with the normal to AC. Because here the incident angle is more than critical angle so rays suffer total internal reflection and reflect at angle of 45°. The beam then strikes face BC, where it is incident normally and so passes undeviated. As a result the incident beam gets deviated through 90°.
Such a prism is used in periscope.

Solution 18.

(a) The angle of incidence at the face AC is 45° and angle of incidence at the face BC is 0°.
(b) The ray suffers total internal reflection at the face AC.

Solution 19.

Solution 20.

Solution 21.

Solution 22.

Solution 23.

A total reflecting prism can be used to turn a ray of light by 180°. The following diagram can make it further clear.

Solution 24.

When total internal reflection occurs from a prism, the entire incident light (100%) is reflected back into the denser medium. Whereas in ordinary reflection from a plane mirror, some light is refracted and absorbed so the reflection is partial.

Solution 25.

A total reflecting prism gives us an image much brighter than that obtained by using a plane mirror.

Solution 1 (MCQ).

42°

Solution 2 (MCQ).

Solution 3 (MCQ).

More Resources for Selina Concise Class 10 ICSE Solutions

## Selina Concise Physics Class 10 ICSE Solutions Refraction through Lens

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens. You can download the Selina Concise Physics ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Physics for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

Selina ICSE Solutions for Class 10 Physics Chapter 5 Refraction through Lens

Exercise 5(A)

Solution 1.

A lens is a transparent refracting medium bounded by two curved surfaces which are generally spherical.

Solution 2.

Solution 3.

Convex lens:

1. It converge the incident rays towards the principal axis.
2. It has a real focus.

Concave lens:

1. It diverges the incident rays away from the principal axis.
2. It has a virtual focus.

Solution 4.

Equiconvex lens is converging.

Solution 5.

Concave lens will show the divergent action on a light beam.

Solution 6.

As shown in the figure the convex lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 7.

As shown in the figure the concave lens has two glass prisms and one glass block. One of the glass prisms is situated above the glass block and one below the block.

Solution 8.

If a parallel beam of light is incident on a convex lens then the upper part of the lens bends the incident ray downwards. The lower part bens the ray upwards while the central part passes the ray undeviated.

But in case of a concave lens the upper part of the lens bends the incident ray upwards and lower part bends the ray downwards while the central part passes the ray undeviated.

Solution 9.

It is the line joining the centers of curvature of the two surfaces of the lens.

Solution 10.

It is point on the principal axis of the lens such that a ray of light passing through this point emerges parallel to its direction of incidence.
It is marked by letter O in the figure. The optical centre is thus the centre of the lens.

Solution 11.

Solution 12.

A lens is called an equiconvex or equiconcave when radii of curvature of the two surfaces of lens are equal.

Solution 13.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a convex lens, the first focal point is a point F1 on the principal axis of the lens such that the rays of light starting from it or passing through it, after refraction through lens, become parallel to the principal axis of the lens.

The second focal point for a convex lens is a point F2 on the principal axis such that the rays of light incident parallel to the principal axis, after refraction from the lens, pass through it.

Solution 14.

A light ray can pass through a lens from either direction. Therefore, a lens has two principal foci.

For a concave lens, the first focal point is a point F1 on the principal axis of the lens such that the incident rays of light appearing to meet at it, after refraction from the lens become parallel to the principal axis of the lens.

The second focal point for a concave lens is a point F2 on the principal axis of the lens such that the rays of light incident parallel to the principal axis, after refraction from the lens, appear to be diverging from this point.

Solution 15.

Solution 16.

Solution 17.

Solution 18.

Solution 19.

Solution 20.

Solution 21.

The distance from the optical centre O of the lens to its second focal point is called the focal length of the lens.

Solution 22.

A plane passing through the focal point and normal to the principal axis of the lens is called the first focal plane.

Solution 23.

(i) If a lens has both its focal length equal medium is same on either side of lens.
(ii)If a ray passes undeviated through the lens it is incident at the optical centre of the lens.

Solution 24.

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Solution 29.

(a) If half part of a convex lens is covered, the focal length does not change, but the intensity of image decreases.
(b) A convex lens is placed in water. Its focal length will increase.
(c) The focal length of a thin convex lens is more than that of a thick convex lens.

Solution 1 (MCQ).

First focus

Solution 2 (MCQ).

Its second focus

Exercise 5(B)

Solution 1.

1. A ray of light incident at the optical centre O of the lens passes undeviated through the lens.
2. A ray of light incident parallel to the principal axis of the lens, after refraction passes through the second focus F2 (in a convex lens) or appears to come from the second focus F2 (in a concave lens).
3. A ray of light passing through the first focus F1 (in a convex lens) or directed towards the first focus F1 (in a concave lens), emerges parallel to the principal axis after refraction.

Solution 2.

Solution 3.

 Real image Virtual image 1. A real image is formed due to actual intersection of refracted (or reflected) rays. 1. A virtual image is formed when the refracted (or reflected) rays meet if they are produced backwards. 2. A real image can be obtained on a screen. 2. A virtual image can not be obtained on a screen. 3. A real image is inverted with respect to the object. 3. A virtual image is erect with respect to the object.

Solution 4.

Solution 5.

Solution 6.

Solution 7.

(ii) The position of the images will be more than twice the focal length of lens.
(iii) The image will be magnified, real and inverted.
(iv) As the object move towards F1 the image will shift away from F2 and it is magnified. At Fthe image will form at infinity and it is highly magnified. Between F1 and optical centre, the image will form on the same side of object and will be magnified.

Solution 8.

Solution 9.

Solution 10.

Solution 13.

Let the candle is placed beyond 2F1 and its diminished image which is real and inverted is formed between F2 and 2F2.

Here the candle is AB and its real and inverted image is formed between F2 and 2F2.

Solution 14.

Solution 15.

Solution 16.

Solution 17.

The object is placed between focal point F1 and convex lens and its image is formed at the same side of the lens which is enlarged.
So this lens can be used as a magnifying lens.

Solution 18.

The sun is at infinity so convex lens forms its image at second focal point which is real and very much diminished in size.

While using the convex lens as burning glass, the rays of light from the sun (at infinity) are brought to focus on a piece of paper kept at the second focal plane of the lens. Due to sufficient heat of the sun rays, the paper burns. Hence this lens is termed as ‘burning glass’.

Solution 19.

(a) This is convex lens.
(b) The nature of the image is real.

Solution 20.

(a) Convex lens.
(b) Virtual.

Solution 21.

(a) Concave lens
(b) Image is diminished

Solution 23.

Image formed by a concave lens is virtual and diminished.

Solution 24.

The virtual image formed by a convex lens will be magnified and upright.

Solution 25.

(a) at focus,
(b) at 2F,
(c) between F and 2F,
(d) between optical centre and focus.

Solution 26.

 Type of lens Position of object Nature of image Size of image Convex Between optic centre and focus Virtual and upright Magnified Convex At focus Real and inverted Very much magnified Concave At infinity Virtual and upright Highly diminished Concave At any distance Virtual and upright Diminished

Solution 27.

1. When the object is situated at infinity, the position of image is at F2, it is very much diminished in size and it is real and inverted.
2. When the object (AB) is situated beyond 2F1, the position of image (A’B’) is between F2 and 2F2, it is diminished in size and real and inverted.
3. When the object (AB) is situated at 2F1, the position of image (A’B’) is at 2F2, it is of same size as the object and real and inverted.
4. When the object (AB) is situated between 2F1and F1, the position of image (A’B’) is beyond 2F2, it is magnified in size and real and inverted.
5. When the object (AB) is situated at F1, the position of image is at infinity; it is very much magnified in size and real and inverted.
6. When the object (AB) is situated between lens and F1, the position of image (CD) is on the same side, behind the object; it is magnified in size and virtual and upright.

Solution 28.

1. When object (AB) is situated at infinity then parallel rays from object appears to fall on concave lens. Due to which image forms at focus. This image is highly diminished in size and virtual and upright.
2. When object (AB) is situated at any point between infinity and optical centre of the lens then image forms between focus and optical centre. This image is diminished in size and virtual and upright.

Solution 29.

(a) An object is placed at a distance of more than 40 cm from a convex lens of focal length 20 cm. The image formed is real, inverted and diminished.
(b) An object is placed at a distance 2f from a convex lens of focal length f. The image formed is equal to that of the object.
(c) An object is placed at a distance 5 cm from a convex lens of focal length 10 cm. The image formed is virtual, upright and magnified.

Solution 30.

(a) False
(b) False
(c) False
(d) True
(e) False

Solution 1 (MCQ).

The focal length of the convex lens is 10 cm.
Hint: As the object distance = image distance, the object must be kept at 2f.
Therefore, 2f = 20 cm or f = 10 cm.

Solution 2 (MCQ).

Virtual and enlarged.
Explanation: When the object is kept between optical centre and focus of a convex lens, the image is formed on the same side, behind the object. The image thus formed is virtual, enlarged and erect.

Solution 3 (MCQ).

Virtual, upright and diminished
Hint: Concave lens forms virtual, upright and diminished image for all positions of the object.

Exercise 5(C)

Solution 1.

1. The axis along which the distances are measured is called as the principal axis. These distances are measured from the optical centre of the lens.
2. All the distances which are measured along the direction of the incident ray of the light are taken positive, while the distances opposite to the direction of the incident ray are taken as negative.
3. All the lengths that are measured above the principal axis are taken positive, while the length below the principal axis is considered negative.
4. The focal length of the convex lens is taken positive and that of concave lens is negative.

Solution 1 (MCQ).

Magnification is -0.5. The negative sign of magnification indicates that the image is real while 0.5 indicates that the image is diminished. A convex lens only forms a real and diminished image of an object. Hence, the correct answer is option (d).

Solution 2.

(i) The positive focal length of a lens indicates that it is a convex lens.
(ii) The negative focal length of a lens indicates that it is a concave lens.

Solution 3.

Lens formula:

• The distance of the object from the optical centre is called the object distance (u).
• The distance of the image from the optical centre is called the image distance (v).
• The distance of the principal focus from the optical centre is called the focal length (f).

Solution 3 (MCQ).

Solution 4.

The term magnification means a comparison between the size of the image formed by a lens with respect to the size of the object.
For a lens: Magnification ‘m’ is the ratio of the height of the image to the height of the object.

Solution 4 MCQ.

Power of a lens is +1.0 D. The positive sign indicates that the focal length of the lens is positive which indicates the lens is a convex lens.

Solution 5.

(i) Positive sign of magnification indicates that the image is virtual while negative sign indicates that the image is real.
(ii) Positive sign of magnification indicates that the image is erect while negative sign indicates that the image is inverted.

Solution 6.

The power of a lens is a measure of deviation produced by it in the path of rays refracted through it.
Its unit is Dioptre (D).

Solution 7.

Solution 8.

If focal length of a lens doubled then its power gets halved.

Solution 9.

The sign of power depends on the direction in which a light ray is deviated by the lens. The power could be positive or negative. If a lens deviates a ray towards its centre (converges), the power is positive and if it deviates the ray away from its centre (diverges), the power is negative.

Solution 10.

It is a concave.

Solution 9 (MCQ).

Solution 10 (MCQ).

Solution 1 (Num).

Solution 2 (Num).

Solution 3 (Num).

Solution 4 (Num).

Solution 5 (Num).

Solution 6 (Num).

Solution 7 (Num).

Solution 8 (Num).

Solution 9 (Num).

Solution 10 (Num).

Solution 11 (Num).

Solution 11 (Num).

Solution 13 (Num).

Exercise 5(D)

Solution 1.

Magnifying glass is a convex lens of short focal length. It is mounted in a lens holder for practical use.
It is used to see and read the small letters and figures. It is used by watch makers to see the small parts and screws of the watch.

Solution 2.

Let the object (AB) is situated between focal length and optical centre of a convex lens then its image (A’B’) will form on the same side of lens.

The image formed will be virtual, magnified and erect.

Solution 3.

The object is placed between the lens and principal focus.
The image is obtained between the lens and principal focus.

Solution 4.

The magnifying power of the microscope is defined as the ratio of the angle subtended by the image at the eye to the angle subtended by the object (assumed to be placed at the least distance of distinct vision D = 25 cm) at the eye, i.e.,

where F is the focal length of the lens.
The magnifying power of a microscope can be increased by using the lens of short focal length. But it cannot be increased indefinitely.

Solution 5.

The two applications of a convex lens are:-

1. It is used as an objective lens in a telescope, camera, slide projector, etc.
2. With its short focal length it is also used as a magnifying glass.

The two applications of a concave lens are:-

1. A person suffering from short sightedness or myopia wears spectacles having concave lens.
2. A concave lens is used as eye lens in a Galilean telescope to obtain an erect final image of the object.

Solution 6.

The approximate focal length of a convex lens can be determined by using the principle that a beam of parallel rays incident from a distant object is converged in the focal plane of the lens.
In an open space, against a white wall, a metre scale is placed horizontally with its 0 cm end touching the wall.

By moving the convex lens to and fro along the scale, focus a distant object on wall. The image which forms on the wall is very near to the focus of the lens and the distance of the lens from the image is read directly by the metre scale. This gives the approximate focal length of the lens.

Solution 7.

Solution 8.

To determine focal length by using plane mirror we need a vertical stand, a plane mirror, a lens and a pin.
Place the lens L on a plane mirror MM’ horizontally. Arrange a pin P on the clamp of a vertical stand such that the tip of pin is vertically above the centre O of the lens.

Adjust the height of the pin until it has no parallax (i.e., when the pin and its image shift together) with its inverted image as seen from vertically above the pin.
Now measure the distance x of the pin from the lens and the distance y of the pin from the mirror, using a metre scale and a plumb line. Calculate the average of the two distances. This gives the focal length of the lens, i.e.,

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