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Physics MCQs for NEET — Practice Questions with Answers

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Match the corresponding entries of Column 1 with Column 2. [Where m is the magnification produced by the mirror]

Column 1 Column 2
A. m=-2 a. Convex mirror
B. m=-1/2 b. Concave mirror
C. m=+2 c. Real image
D. m=+1/2 d. Virtual Image

A->a and c;B->a and d; C->a and b; D->c and d

A->a and d; B->b and c; C->b and d; D-> b and c

A->c and d; B->b and d;C->b and c;D->a and d

A->b and c; B->b and c; C->b and d; D->a and d

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Explanation

(d) A concave minor forms real and virtual images, whose magnification can be negative or positive depending upon the position of the object. If object is placed between focus and pole the image obtained will be virtual and its magnification will be positive. In all other cases concave mirror forms real images whose magnification will be negative. A convex mirror always forms a virtual image whose magnification will always be positive.

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The angle of incidence for a ray of light at a refracting surface of a prism is 45°. The angle of prism is 60°. If the ray suffers minimum deviation through the prism, the angle of deviation and refracting index of the material of the prism respectively are

30°, $\sqrt{2}$

45°, $\sqrt{2}$

30°, $\frac{1}{\sqrt{2}}$

45°, $\frac{1}{\sqrt{2}}$

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In an astronomical telescope in normal adjustment a straight black line of length L is drawn on inside part of the objective lens. The eye-piece forms a real image of this line. The length of this image is l. The magnification of the telescope is:

L/l+1

L/l-1

L+1/L-1

L/l

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Explanation

In an astronomical telescope, the magnification is given by the ratio of the angle subtended by the image at the eye to the angle subtended by the object at the objective lens. This is equal to the ratio of the focal length of the objective lens to the focal length of the eyepiece. Therefore, the magnification is simply the ratio of the length of the object (L) to the length of the image (l).

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

microscope will increase but that of telescope decrease

microscope and telescope both will increase

microscope and telescope both will decrease

microscope will decrease but that of telescope will increase

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Explanation

$For microscope, m = \frac{L}{f_{o}} \times \frac{D}{f_{e}} \Rightarrow m \propto \frac{1}{f_{o}} For telescope, m = \frac{f_{o}}{f_{e}} \Rightarrow f \propto f_{o}$

The magnifying power of the microscope will decrease but the magnifying power telescope will increase.

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The angle of a prism is A. One of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2A on the first surface returns back through the same path after suffering reflection at the silvered surface. The refractive index μ of the prism is

2sinA

2cosA

1/2cosA

tanA

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Explanation

When a light ray falls on a prism at an angle A and gets reflected from the silvered surface, the angle of incidence is 2A. For this to happen, the refractive index of the prism should be 2cosA, according to Snell's law.

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A plane-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices in μ₁ and μ₂ and R is the radius of curvature of the curved surface of the lenses. then the focal length of the combination is

R/2(μ₁₊μ₂₎

R/2(μ_1-μ₂₎

R/(μ_1-μ₂₎

2R/(μ2-μ₁₎

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Explanation

(c) Focal length of the combination

1/f=1/f₁+1/f_{2 ...(i)}

We have f₁=R/(μ_1-1) and f₂=R/(μ₂-1)

or 1/f₁=R/(μ_1-1) or 1/f₂=-R/(μ₂-1)

Putting these values in Eq. (i), we get

1/f=(μ_1-1)/R-(μ₂-1)/R

=[μ_1-1-μ₂₊₁)/R]=(μ_1-μ_2/R)

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

5 cm

25 cm

1.67 cm

1.5 cm

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Explanation

(c)

$G i v e n : P_{1} = 40 D P_{2} = 20 D P_{n e t} = 40 + 20 = 60 D F o c a l l e n g t h = \frac{1}{P} = \frac{1}{60} m = \frac{100}{60} = 1.67 c m$

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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

equal to that of glass

less than one

greater than that of glass

less than that of glass

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Explanation

If biconvex lens behave like plane sheet, ray will pass undeviated through it only when medium has same refractive index as that of biconvex lens.

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A concave mirror of focal length $f_{1}$ is placed at a distance of d from a convex lens of focal length $f_{2} .$ A beam of light coming from infinity and falling on this convex lens concave mirror combination returns to infinity. The distance d must be equal

$f_{1} + f_{2}$

$- f_{1} + f_{2}$

$2 f_{1} + f_{2}$

$- 2 f_{1} + f_{2}$

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

10cm,10cm

15cm,5cm

18cm,2cm

11cm,9cm

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Explanation

Given, $M = \frac{f_{0}}{f_{e}} = 9 a n d f_{0} + f_{e} = 20$

$f_{0} = 9 f_{e}$

So, $9 f_{e} + f_{e} = 20$

$f_{e} = 2 c m$

$f_{0} = 9 \times 2$

$f_{0} = 18 c m$

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