Question 1

An image formed by a plane mirror is:

Accepted Answer

erect and of same size as the object. A plane mirror forms a virtual image that is upright (erect) and has the same size as the object. It does not magnify or diminish the object, so the image and object are equal in size.

Question 2

One face of a rectangular glass plate 6 cm thick is silvered. An object held 8 cm in front of the first face forms an image 12 cm behind the silvered face. The refractive index of the glass is

Accepted Answer

1.6. The apparent shift due to refraction is given by the formula: shift = t(1 - 1/μ), where t is the thickness of the glass and μ is the refractive index. The total distance between the object and the image is 8 cm (object distance) + 6 cm (thickness of glass) + 12 cm (distance behind the silvered face) = 26 cm. The apparent distance through the glass is 6/μ. Solving for μ using the given distances, we find μ = 1.6.

Question 3

Time taken by sunlight to pass through a window of thickness 4 mm whose refractive index is 3/2 is

Accepted Answer

2 × 10⁻⁴ sec. The time taken by light to pass through a medium is given by t = d / v, where d is the thickness and v is the speed of light in the medium. The speed of light in the medium is v = c / μ, where c is the speed of light in a vacuum (3 × 10⁸ m/s) and μ is the refractive index. Substituting μ = 3/2 and d = 4 mm = 4 × 10⁻³ m, we get t = (4 × 10⁻³) / (3 × 10⁸ / (3/2)) = 2 × 10⁻⁴ sec.

Question 4

A beam of monochromatic light is refracted from vacuum into a medium of refractive index 1.5, the wavelength of refracted light will be

Accepted Answer

smaller. When light enters a medium with a refractive index greater than 1, its speed decreases, and consequently, its wavelength becomes smaller. The frequency remains unchanged.

Question 5

Green light of wavelength 5460 Å is incident on an air-glass interface. If the refractive index of glass is 1.5, the wavelength of light in glass would be (c = 3 × 10⁸ ms⁻¹)

Accepted Answer

3640 Å. The wavelength of light in a medium is given by λ_medium = λ_air / μ, where μ is the refractive index. Substituting λ_air = 5460 Å and μ = 1.5, we get λ_glass = 5460 / 1.5 = 3640 Å.

Question 6

When a biconvex lens of glass having refractive index 1.47 is dipped in a liquid, it acts as a plane sheet of glass. This implies that the liquid must have refractive index.

Accepted Answer

equal to that of glass. For the lens to act as a plane sheet of glass, there should be no refraction at its surfaces. This happens when the refractive index of the liquid is equal to that of the glass, as the refractive index difference becomes zero.

Question 7

A concave mirror of focal length 'f₁' is placed at a distance of 'd' from a convex lens of focal length 'f₂'. A beam of light coming from infinity and falling on this convex lens-concave mirror combination returns to infinity. The distance 'd' must equal:

Accepted Answer

2f₁ + f₂. For the light to return to infinity, the system must act as an equivalent plane mirror. This happens when the distance between the lens and the mirror equals the sum of twice the focal length of the concave mirror (2f₁) and the focal length of the convex lens (f₂).

Question 8

A biconvex lens has a radius of curvature of magnitude 20 cm. Which one of the following options best describe the image formed of an object of height 2 cm placed 30 cm from the lens?

Accepted Answer

Real, inverted, height = 1 cm. Using the lens maker's formula and magnification formula, the lens is converging with a focal length of 10 cm. For an object at 30 cm, the image is real, inverted, and reduced in size with a height of 1 cm.

Question 9

If the focal length of objective lens is increased then magnifying power of:

Accepted Answer

microscope will decrease but that of telescope increase.. For a microscope, increasing the focal length of the objective lens reduces its magnifying power as magnification is inversely proportional to the focal length. For a telescope, increasing the focal length of the objective lens increases its magnifying power as magnification is directly proportional to the focal length of the objective lens.

Question 10

For a normal eye, the cornea of eye provides a converging power of 40D and the least converging power of the eye lens behind the cornea is 20D. Using this information, the distance between the retina and the eye lens of the eye can be estimated to be

Accepted Answer

1.67 cm. The total power of the eye is the sum of the cornea and lens powers, i.e., 40D + 20D = 60D. Using the lens formula, focal length f = 1/P, where P is in diopters, we get f = 1/60 m = 0.0167 m = 1.67 cm. This corresponds to the distance between the retina and the eye lens.

Question 11

The magnifying power of a telescope is 9. When it is adjusted for parallel rays the distance between the objective and eyepiece is 20 cm. The focal length of lenses are:

Accepted Answer

15 cm, 5 cm. The magnifying power of a telescope for parallel rays is given by M = f_o / f_e, where f_o and f_e are the focal lengths of the objective and eyepiece, respectively. Also, the distance between the lenses is f_o + f_e. Solving the equations M = 9 and f_o + f_e = 20 cm, we get f_o = 15 cm and f_e = 5 cm.

Question 12

A point source of light is placed 4 m below the surface of water of refractive index 5/3. The minimum diameter of a disc, which should be placed over the source, on the surface of water to cut off all light coming out of water is

Accepted Answer

3 m. The light rays from the source will undergo total internal reflection at the critical angle. The radius of the circle of light on the water surface is determined by the critical angle and the depth of the source. Using Snell's law and geometry, the minimum diameter of the disc is 3 m.

Question 13

An astronomical telescope has a length of 44 cm and tenfold magnification. The focal length of the objective lens is

Accepted Answer

40 cm. The length of an astronomical telescope is given by the sum of the focal lengths of the objective and eyepiece lenses. For a magnification of 10, the focal length of the objective lens is 10 times that of the eyepiece. Let the focal length of the eyepiece be f_e. Then, f_o = 10f_e. From the length equation, f_o + f_e = 44 cm. Substituting f_o = 10f_e, we get 10f_e + f_e = 44, or 11f_e = 44, so f_e = 4 cm and f_o = 40 cm.

Question 14

A microscope is focussed on a mark on a piece of paper and then a slab of glass of thickness 3 cm and refractive index 1.5 is placed over the mark. How should the microscope be moved to get the mark in focus again?

Accepted Answer

1 cm upward. When a glass slab of thickness t and refractive index μ is placed, the apparent shift of the object is given by t(1 - 1/μ). Substituting t = 3 cm and μ = 1.5, the shift is 3(1 - 1/1.5) = 1 cm. The microscope must be moved 1 cm upward to refocus.

Question 15

A transparent solid is invisible in a liquid of same refractive index because of no refraction.

Accepted Answer

1/f = 0. When the refractive index of the solid (μg) matches that of the liquid (μm), there is no refraction at the interface, leading to no focusing effect, and the focal length becomes infinite (1/f = 0).

Question 16

The hypermetropia is a

Accepted Answer

long-sight defect. Hypermetropia, also known as farsightedness or long-sight defect, is a condition where a person can see distant objects clearly but has difficulty focusing on nearby objects.

Question 17

A lens system has R = 20 cm, h0 = 2 cm, and u = -30 cm. Find the magnification and the nature of the image.

Accepted Answer

hi = v/u × h0 = -4 cm. The magnification formula is m = hi/h0 = v/u, and the image height hi can be calculated as hi = (v/u) × h0. Substituting the given values, hi = (-60/-30) × 2 = -4 cm. This indicates the image is inverted.

Question 18

A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move thr

Accepted Answer

y / 2x. When the mirror is rotated by an angle θ, the reflected beam rotates by 2θ. Using the geometry of the situation, the displacement y on the scale is related to the angle of rotation by the formula θ = y / 2x, where x is the distance between the mirror and the scale.

Question 19

Match the corresponding entries of column-1 with column-2 (Where m is the magnification produced by the mirror):
Column-1:
(P) m = −2
(Q) m = −1/2
Column-2:
(A) Convex mirror
(B) Concave mirror

Accepted Answer

P → C and D, Q → B and D, R → B and C, S → A and D. For P (m = -2), the negative sign indicates an inverted image, which is possible with a concave mirror (C) when the object is between the center of curvature and the focus. For Q (m = -1/2), the negative sign again indicates an inverted image, which is possible with a concave mirror (D) when the object is beyond the center of curvature. Thus, the correct matching is P → C and D, Q → B and D.

Question 20

A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that its end closer to the pole is 20 cm away from the mirror. The length of the image is:

Accepted Answer

5 cm. The image of the rod is formed by considering the magnification at each end. For the end at 20 cm, magnification is -1/2, and for the end at 30 cm, magnification is -1/3. The difference in image positions gives the length of the image as 5 cm.

NEET Physics: Ray Optics and Optical Instruments questions with solutions

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