Question 1

A telescope is fitted with an objective lens of focal length 100 cm and an eyepiece of focal length 5 cm. If the final image is adjusted to form at the least distance of distinct vision, what is the telescope’s magnifying power?

Accepted Answer

24. For final image at the least distance of distinct vision (D = 25 cm), M = (fo/fe)(1 + fe/D) = (100/5)(1 + 5/25) = 20 x 1.2 = 24.

Question 2

An object placed 2.4 m in front of a lens produces a clear image on a film located 12 cm behind the lens. A 1 cm thick glass slab of refractive index 1.50 is then inserted between the lens and the film, with its plane surfaces parallel to the film. To obtain a sharp image again on the film

Accepted Answer

5.6 m. The insertion of the glass slab increases the optical path length due to its refractive index, requiring the object to be moved closer to the lens to maintain a sharp image on the film. The new object distance is calculated based on the effective change in distance caused by the slab, resulting in the correct answer of 5.6 m.

Question 3

A light ray is incident on an equilateral prism in such a way that the angle of incidence equals the angle of emergence, and this common angle is three-fourths of the prism angle. What is the angle of deviation produced?

Accepted Answer

30°. Equilateral prism A=60 deg, i=e=(3/4)(60)=45 deg. Deviation delta = i + e - A = 45 + 45 - 60 = 30 deg.

Question 4

A glass lens with refractive index 1.60 has a focal length of +20 cm when placed in air. If the same lens is immersed in water of refractive index 1.33, its focal length will be

Accepted Answer

three times the air value. f_water/f_air = (n_g - 1)/((n_g/n_w) - 1) = (1.60 - 1)/((1.60/1.33) - 1) = 0.60/0.203 = 2.95, i.e. about three times the air focal length.

Question 5

A prism is placed in air. If a ray is incident at 60° and the prism angle, the deviation angle δ, and the angle of emergence e are all equal, what is the refractive index of the prism?

Accepted Answer

1.73. With angle of incidence = emergence = 60 deg and A = delta = e, the prism is in symmetric (minimum-deviation) configuration with A = 60 deg, delta = 60 deg. n = sin((A+delta)/2)/sin(A/2) = sin60/sin30 = 0.866/0.5 = 1.73.

Question 6

A person’s farthest point of clear vision is 30 cm from his eyes. If he wants to read a book held 50 cm away, what power of lens should he use?

Accepted Answer

−1.33 D. The lens must image an object at 50 cm to the far point at 30 cm: 1/v - 1/u = 1/f with u = -50 cm, v = -30 cm gives 1/f = -1/30 + 1/50 = -1/75 cm -> f = -75 cm. P = 100/(-75) = -1.33 D.

Question 7

An object is kept 40 cm in front of a concave mirror whose focal length is 20 cm. The image formed will be

Accepted Answer

real, inverted and equal in size. With u = 40 cm and f = 20 cm, the object sits at the centre of curvature (2f). The image also forms at 2f: it is real, inverted and equal in size to the object.

Question 8

Two plano-convex lenses, labelled 1 and 2, are made of glass with refractive index 1.5. Their radii of curvature are 25 cm and 20 cm, respectively. They are kept in contact with their curved faces facing one another, and the gap between them is occupied by a liquid of refractive index 4/3.

Accepted Answer

a convex lens with focal length 66.6 cm. f1 = 25/0.5 = 50 cm (P=0.02), f2 = 20/0.5 = 40 cm (P=0.025). The 4/3 liquid forms a biconcave lens: 1/f = (1/3)(1/(-25) - 1/20) = -0.03. Total P = 0.02 + 0.025 - 0.03 = 0.015, so f = +66.6 cm, a convex lens.

Question 9

A microscope is adjusted to focus on a spot marked on a sheet of paper. A glass slab 3 cm thick with refractive index 1.5 is then placed over the mark. In what direction and by how much must the microscope be shifted to bring the mark back into focus?

Accepted Answer

1 cm upward. The slab raises the apparent position by t(1 - 1/n) = 3(1 - 1/1.5) = 3(1/3) = 1 cm. The microscope must be moved 1 cm upward to refocus.

Question 10

A transparent solid sphere has refractive index 1.5 and contains a tiny mark at its centre. If the sphere is viewed from outside, where will the mark appear to be located?

Accepted Answer

At the same position as its actual location. The mark appears at the same position as its actual location because the light rays from the mark travel through the sphere and do not bend significantly at the center, resulting in no apparent shift in position when viewed from outside.

Question 11

For a spherical convex mirror, how is the focal length f connected to its radius of curvature r?

Accepted Answer

f = +r/2. The mirror relation is f = R/2. In the Cartesian sign convention the center of curvature of a convex mirror is behind it, so R is positive and the focal length is positive: f = +r/2.

Question 12

Why is an astronomical telescope made with a large aperture?

Accepted Answer

To obtain greater resolving power. A larger aperture allows more light to enter the telescope, which enhances its ability to distinguish fine details and separate closely spaced objects, thereby increasing its resolving power.

Question 13

When two plane mirrors are placed so that the angle between them is 60°, how many images of an object are produced?

Accepted Answer

5. When two mirrors are positioned at an angle of 60°, the number of images formed can be calculated using the formula (360°/angle) - 1. In this case, (360°/60°) - 1 equals 6 - 1, resulting in 5 images.

Question 14

Which optical principle is employed for the transmission of light through optical fibres?

Accepted Answer

Total internal reflection. Total internal reflection is the principle that allows light to be transmitted through optical fibers by reflecting light within the fiber core, ensuring minimal loss of signal as it travels along the length of the fiber.

Question 15

In a compound microscope, the objective lens produces an image that is

Accepted Answer

real and magnified. The objective lens in a compound microscope creates a real image that is larger than the object being observed, allowing for detailed examination of small specimens.

Question 16

At what angle should two plane mirrors be placed so that a single object forms three images?

Accepted Answer

90°. When two plane mirrors are positioned at a 90° angle to each other, they create multiple reflections that result in three distinct images of the object. This occurs because each mirror reflects the image created by the other, leading to a total of three visible images.

Question 17

A plano-convex lens has refractive index 1.5 and the radius of curvature of its curved face is 30 cm. The curved face is silvered, and the lens is then used to obtain an image of an object. At what distance from the lens should the object be placed so that the image formed is real and equa

Accepted Answer

20 cm. The object must be placed at a distance equal to twice the focal length of the lens to produce a real image that is the same size as the object. Given the radius of curvature and the refractive index, the focal length can be calculated, leading to the conclusion that the object should be placed 20 cm from the lens.

Question 18

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12 cm below the surface, the radius of this circle in cm is

Accepted Answer

36/√7. Critical angle: sin(theta_c)=1/n=3/4, so tan(theta_c)=3/sqrt(7). Radius r = h*tan(theta_c) = 12*3/sqrt(7) = 36/sqrt(7) cm.

Question 19

A thin glass (refractive index 1.5) lens has optical power of −5 D in air. Its optical power in a liquid medium with refractive index 1.6 will be

Accepted Answer

1 D. P_liquid = P_air * (n_g/n_l - 1)/(n_g - 1) = -5 * (1.5/1.6 - 1)/(1.5 - 1) = +0.625 D. The lens becomes weakly converging (positive power), nearest to 1 D.

Question 20

The refractive index of a glass is 1.520 for red light and 1.525 for blue light. Let D1 and D2 be the angles of minimum deviation for red and blue respectively in a prism of this glass. Then,

Accepted Answer

D1 < D2. The refractive index of a material determines how much light bends when passing through it; since red light has a lower refractive index than blue light, it will experience less deviation in the prism, resulting in a smaller angle of minimum deviation for red light compared to blue light.

Question 21

Two lenses of power −15 D and +5 D are in contact with each other. The focal length of the combination is

Accepted Answer

−10 cm. The focal length of a lens combination can be calculated using the formula 1/f = 1/f1 + 1/f2. For the given powers, the focal lengths are -6.67 cm and +20 cm, respectively, leading to a combined focal length of -10 cm.

Question 22

An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distances are measured by

Accepted Answer

a vernier scale provided on the microscope. A travelling microscope measures small distances using the vernier scale fitted on it, allowing readings to about 0.01 mm.

Question 23

In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plotted using t

Accepted Answer

(2f, 2f). The coordinates of point P represent the object and image distances where the lens formula (1/f = 1/v + 1/u) holds true. Since the line at 45° indicates that u = v, and given the nature of a convex lens, the point where both distances equal 2f satisfies the lens equation, confirming that (2f, 2f) is the correct answer.

Question 24

Let the x-z plane be the boundary between two transparent media. Medium 1 in z ≥ 0 has a refractive index of √2 and medium 2 with z < 0 has a refractive index of √3. A ray of light in medium 1 given by the vector A = 6√3 i + 8√3 j − 10 k is incident on the plane of separation. The angle of

Accepted Answer

45°. With the boundary at z=0, |A| = 20 and A_z = -10 give cos(theta1) = 1/2, so theta1 = 60 deg and sin(theta1) = sqrt3/2. Snell's law sqrt2 * sqrt3/2 = sqrt3 * sin(theta2) gives sin(theta2) = 1/sqrt2, so theta2 = 45 deg.

Question 25

A plano-convex lens has a diameter of 6 cm and a central thickness of 3 mm. If the speed of light in the lens material is 2 × 10⁸ m/s, what is the focal length of the lens?

Accepted Answer

30 cm. The focal length of a plano-convex lens can be calculated using the lensmaker's equation, which takes into account the curvature of the lens surfaces and the refractive index of the material. Given the parameters, the calculated focal length is 30 cm, which aligns with the properties of the lens and the speed of light in the material.

Question 26

A thin convex lens made of crown glass with refractive index 3/2 has focal length f in air. If the same lens is immersed separately in two liquids of refractive indices 4/3 and 5/3, its focal lengths become f1 and f2, respectively. Which of the following correctly describes the relation am

Accepted Answer

f1 is greater than f and f2 becomes negative. When the lens is immersed in a liquid with a refractive index lower than that of the lens (like 4/3), its focal length increases, making f1 greater than f. However, when immersed in a liquid with a higher refractive index (5/3), the lens behaves as if it is diverging, resulting in a negative focal length for f2.

Question 27

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears:

Accepted Answer

20 times nearer. A telescope of magnifying power 20 increases the angular size, so the tree appears 20 times nearer (not taller); apparent linear height is unchanged.

Question 28

In an experiment for determination of refractive index of a prism by i − δ, plot it was found that a ray incident at angle 35°, suffers a deviation of 40° and that it emerges at angle 79°. In that case which of the following is closest to the maximum possible value of the refractive index?

Accepted Answer

1.5. From delta = i + e - A, the prism angle A = 35 + 79 - 40 = 74 deg. Working through the refraction at the faces, the largest consistent refractive index is about 1.5 (JEE Main 2016).

Question 29

A diverging lens with magnitude of focal length 25 cm is placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm. A beam of parallel light falls on the diverging lens. The final image is formed:

Accepted Answer

real and at a distance of 40 cm from the convergent lens. The correct option is right because the diverging lens first creates a virtual image that acts as the object for the converging lens. The converging lens then produces a real image at a distance of 40 cm from itself, which is consistent with the lens formula and the behavior of light through these two types of lenses.

Question 30

A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. The

Accepted Answer

0.55 cm away from the lens. When the glass block is placed in front of the light source, it effectively increases the optical path length due to its refractive index. This causes the light rays to converge at a point further away from the lens, necessitating a shift of the screen by 0.55 cm to maintain a sharp image.

Question 31

A concave mirror for face viewing has focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is:

Accepted Answer

0.32m. For an erect virtual image m=+5 in a concave mirror (f=-0.4 m): m=-v/u=+5 gives v=-5u. Using 1/v+1/u=1/f with u=-d: -4/(5d) = 1/(-0.4) -> 5d = 1.6 -> d = 0.32 m.

Question 32

If we need a magnification of 375 from a compound microscope of tube length 150 mm and an objective of focal length 5 mm, the focal length of the eye-piece, should be close to:

Accepted Answer

22mm. To achieve the desired total magnification of 375 with a compound microscope, we can use the formula for total magnification, which is the product of the objective magnification and the eyepiece magnification. Given the tube length and the focal length of the objective, we can calculate the required focal length of the eyepiece, which results in approximately 22 mm.

Question 33

Directions: Questions number 10-12 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index μ(I) = μ0 + μ2 I, where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing

Accepted Answer

convex. The initial shape of the wavefront is convex because, in a parallel beam, the wavefronts are typically straight lines that curve outward, resembling a convex shape. This is consistent with the behavior of light traveling in a uniform medium before any refraction occurs.

Question 34

Directions: Questions number 10-12 are based on the following paragraph. An initially parallel cylindrical beam travels in a medium of refractive index μ(I) = μ0 + μ2 I, where μ0 and μ2 are positive constants and I is the intensity of the light beam. The intensity of the beam is decreasing

Accepted Answer

minimum on the axis of the beam. Refractive index mu = mu0 + mu2*I with mu2 > 0, so mu is largest where I is largest, i.e. on the axis. Since v = c/mu, the speed is minimum on the axis of the beam.

Question 35

Q.67 A green light is incident from the water to the air- water interface at the critical angle (θ). Select the correct statement -

Accepted Answer

The spectrum of visible light whose frequency is less than that of green light will come out to the air medium. Refractive index increases with frequency, so higher-frequency light has a smaller critical angle. Green is incident exactly at its own critical angle; colors of higher frequency (smaller critical angle) exceed their critical angle and are totally internally reflected, while colors of lower frequency (larger critical angle) are below theirs and refract out into the air.

Question 36

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears:

Accepted Answer

20 times nearer. A telescope's magnifying power describes angular magnification: the object subtends an angle 20x larger, so the tree appears 20 times nearer (not literally 20 times taller). Answer: 20 times nearer.

Question 37

A ray of light is incident at an angle of 60° on one face of a prism of angle 30°. The emergent ray of light makes an angle of 30° with incident ray. The angle made by emergent ray with second face of prism will be -

Accepted Answer

90°. The emergent ray makes an angle of 30° with the incident ray, and since the prism has an angle of 30°, the light exits at a right angle to the second face, resulting in an angle of 90° between the emergent ray and the second face.

Question 38

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are

Accepted Answer

18 cm, 20 cm. For separated lenses, 1/F = 1/f1 + 1/f2 - d/(f1*f2). Testing f1=18, f2=20 cm with d=2: 1/18 + 1/20 - 2/360 = 0.1000, giving F = 10 cm. So the component focal lengths are 18 cm and 20 cm.

Question 39

The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will

Accepted Answer

3.1 cm. For a single refracting surface with a parallel incident beam, n2/v = (n2-n1)/R, so v = n2*R/(n2-n1) = 1.34*7.8/(0.34) = 30.7 mm = 3.1 cm.

Question 40

A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image

Accepted Answer

10 cm. The correct option is 10 cm because the concave mirror can only produce a virtual image when the object is placed within its focal length. Since the focal length of the concave mirror is equal to the distance from the mirror to the image formed by the lens, which is 10 cm, any object placed beyond this distance will not yield a virtual image.

Question 41

A concave mirror for face viewing has focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is:

Accepted Answer

0.32 m. To see an upright image with a magnification of 5 using a concave mirror, the object distance must be set at a specific point that relates to the focal length and magnification. Using the mirror formula and magnification equation, the calculations show that holding the mirror at 0.32 m achieves the required conditions.

Question 42

A plano-convex lens (focal length f2, refractive index μ2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index μ1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

Accepted Answer

R/(μ2 - μ1). Plano-convex: 1/f2 = (mu2-1)/R; plano-concave: 1/f1 = -(mu1-1)/R. In contact, 1/f = (mu2-mu1)/R, so f = R/(mu2 - mu1).

Question 43

The magnifying power of a telescope with tube length 60 cm is 5. What is the focal length of its eye piece ?

Accepted Answer

10 cm. The magnifying power of a telescope is given by the ratio of the focal lengths of the objective lens to the eyepiece. Given the tube length and magnifying power, we can derive that the focal length of the eyepiece must be 10 cm to achieve the specified magnification.

Question 44

There is a small source of light at some depth below the surface of water (refractive index = 4/3) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly): [Use the fact

Accepted Answer

17 %. The percentage of light emerging from the water surface can be calculated using Snell's law and the critical angle for total internal reflection. Given the refractive index of water, the critical angle is approximately 48.6 degrees, which allows only a fraction of the light to escape, resulting in about 17% of the light emerging from the surface.

Question 45

The refractive index of a converging lens is 1.4. What will be the focal length of this lens if it is placed in a medium of same refractive index ? (Assume the radii of curvature of the faces of lens are R1 and R2 respectively)

Accepted Answer

Infinite. By the lensmaker formula 1/f = (n_lens/n_medium - 1)(1/R1 - 1/R2). When the lens is in a medium of the same refractive index, n_lens/n_medium = 1, so 1/f = 0 and the focal length is infinite.

Question 46

The thickness at the centre of a plano convex lens is 3 mm and the diameter is 6 cm. If the speed of light in the material of the lens is 2 × 10⁸ m s⁻¹, the focal length of the lens is

Accepted Answer

30 cm. The focal length of a plano convex lens can be calculated using the lens maker's formula, which relates the curvature of the lens surfaces and the refractive index of the material. Given the thickness and diameter, the calculations yield a focal length of 30 cm, indicating that the lens has a relatively long focal distance.

Question 47

Your friend is having eye sight problem. She is not able to see clearly a distant uniform window mesh and it appears to her as non-uniform and distorted. The doctor diagnosed the problem as:

Accepted Answer

(1) Astigmatism. When a uniform grid appears non-uniform and distorted (some lines sharper than others), the defect is astigmatism, caused by unequal curvature of the cornea/lens in different meridians.

Question 48

The same size images are formed by a convex lens when the object is placed at 20cm or at 10cm from the lens. The focal length of convex lens is ____ cm.

Accepted Answer

15. The focal length of a convex lens can be determined using the lens formula. When the object distances are at 20 cm and 10 cm, the image distances will be equal due to the same size images formed, allowing us to derive that the focal length is 15 cm.

Question 49

A prism of refractive index μ and angle of prism A is placed in the position of minimum angle of deviation. If minimum angle of deviation is also A, then in terms of refractive index

Accepted Answer

2 cos⁻¹(μ/2). At minimum deviation, mu = sin((A+Dm)/2)/sin(A/2). With Dm = A this gives mu = sin(A)/sin(A/2) = 2cos(A/2), so cos(A/2) = mu/2 and A = 2 cos^-1(mu/2).

Question 50

A short straight object of height 100 cm lies before the central axis of a spherical mirror whose focal length has absolute value |f| = 40 cm. The image of object produced by the mirror is of height 25 cm and has the same orientation as the object. One may conclude from the information:

Accepted Answer

Image is virtual, opposite side of convex mirror.. The image is virtual and located on the opposite side of a convex mirror because convex mirrors always produce virtual images that are upright and smaller than the object, which aligns with the given height and orientation of the image.

JEE Main Physics: Ray Optics and Optical Instruments questions with solutions

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