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Physics NEET Questions — Free MCQ Practice

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A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, is of the order of

A.

0.5 m

B.

5 m

C.

5 mm

D.

5 cm

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Two point white dots are 1mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm. Approximately, what is the maximum distance at which dots can be resolved by the eye ? [Take wavelength of light = 500 nm]

A.

6 m

B.

3 m

C.

5 m

D.

1 m

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Explanation

(c)

$\frac{1.22 λ}{a} = \frac{x}{d} \Rightarrow d = \frac{x \times a}{1.22 λ} = \frac{1 \times 10^{- 3} \times 3 \times 10^{- 3}}{1.22 \times 500 \times 10^{- 9}} = 5 m$

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A room (cubical) is made of mirrors. An insect is moving along the diagonal on the floor such that the velocity of image of insect on two adjacent wall mirrors is $10 c m s^{- 1}$ . The velocity of image of insect in ceiling mirror is

A.

10 $c m s^{- 1}$

B.

20 $c m s^{- 1}$

C.

$\frac{10}{\sqrt{2}} c m s^{- 1}$

D.

$10 \sqrt{2} c m s^{- 1}$

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A thin prism having refracting angle $10^{°}$ is made of glass of refractive index 1.42. This prism is combined with another thin prism of glass of refractive index 1.7. This combination produces dispersion without deviation. The refracting angle of second prism should be

A.

$4^{°}$

B.

$6^{°}$

C.

$8^{°}$

D.

$10^{°}$

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Explanation

(b) For dipersion without deviation, net deviation, produced by the combination of prisms must be zero.

Let, prism angle of the first and second prisms are $A_{1} a n d A_{2}$ respectively. Similarly, their refractive indicates are $μ_{1} a n d μ_{2}$ .

Conditions for dispersion without deviation is

$δ_{1} - δ_{2} = 0$

$\Rightarrow (μ_{1} - 1) A_{1} - (μ_{2} - 1) A_{2} = 0$

$\Rightarrow A_{2} = (\frac{μ_{1} - 1}{μ_{2} - 1}) A_{1} = (\frac{1.42 - 1}{1.7 - 1}) (10^{°})$

$\Rightarrow A_{2} = 6^{°}$

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A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x 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 light is found to move through a distance y on the scale. The angle $θ$ is given by

A.

$\frac{y}{2 x}$

B.

$\frac{y}{x}$

C.

$\frac{x}{2 y}$

D.

$\frac{x}{y}$

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The ratio of resolving powers of an optical microscope for two wavelengths $λ_{1} = 4000 \overset{o}{A} a n d λ_{2} = 6000 \overset{o}{A}$ is

A.

8:27

B.

9:4

C.

3:2

D.

16:81

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Explanation

(c) As, resolving power of a microscope,

$(R P) \propto \frac{1}{λ_{(w a v e l e n g h t)}} ∴ \frac{R P_{1}}{R P_{2}} = \frac{λ_{1}}{λ_{2}} = \frac{6000}{4000} = \frac{3}{2}$

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Two identical glass $(μ_{g} = 3 / 2)$ equi-convex lenses of focal length $f$ each are kept in contact. The space between the two lenses is filled with water $(μ_{w} = 4 / 3)$ . The focal length of the combination is

A.

$f / 3$

B.

$f$

C.

$\frac{4 f}{3}$

D.

$\frac{3 f}{4}$

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An air bubble in a glass slab with refractive index 1.5 (near normal incidence) is 5 cm deep when viewed from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is

A.

8

B.

10

C.

12

D.

16

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Explanation

Let thickness of the given slab ist.

Refractive index of glass = $\frac{R e a l d e p t h}{A p p a r e n t d e p t h}$

Reak depth = n $\times$ apparent depth

When viewed from one side:

Real depth, $d_{1} = 1.5 \times 5 = 7.5 c m$

When viewed from other side:

Real depth, $d_{2} = 1.5 \times 3 = 4.5 c m$

Thick of the slab = 7.5 + 4.5 = 12 cm

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A person can see clearly objects only when they lie between 50 cm and 400 cm from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens, the person has to use will be-

A.

convex, +2.25 diopter

B.

concave, - 0.25 diopter

C.

concave, - 0.2 diopter

D.

convex, + 0.15 diopter

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Explanation

(b) image of object at $\infty$ must lie within distance upto which person can view clearly.

Image distance, v=400cm =4m $\Rightarrow$ u= $\infty$ using lens equation,

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{- 4} - \frac{1}{\infty} = \frac{1}{f} \Rightarrow f = - 4 m$

Now power of the required lens is,

$P = \frac{1}{f} = \frac{1}{- 4} = - 0.25 D$

Thus the person require a concave lens of power -0.25 D

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An astronomical telescope has an objective and eyepiece of focal lengths 40 cm and 4 cm respectively. To view an object 200 cm away from the objective, the lenses must be separated by a distance of :

A.

46.0 cm

B.

50.0 cm

C.

54.0 cm

D.

37.3 cm

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Physics MCQs for NEET — Practice Questions with Answers

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