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NEET NEET Questions — Free MCQ Practice

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A plano-convex lens when silvered in the plane side behaves like a concave mirror of focal length 30 cm. However, when silvered on the convex side it behaves like a concave mirror of focal length 10 cm. Then the refractive index of its material will be

A.

3.0

B.

2.0

C.

2.5

D.

1.5

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Explanation

(d) Here $\frac{1}{F} = - \frac{2}{f} + \frac{1}{f_{m}}$
Plano-convex lens silvered on plane side has $f_{m} = \infty$ .

$∴ \frac{1}{F} = - \frac{2}{f} + \frac{1}{\infty} \Rightarrow \frac{1}{30} = \frac{2}{f} \Rightarrow f = 60 c m$

Plano-convex lens silvered on convex side has $f_{m} = \frac{- R}{2}$

$∴ \frac{1}{F} = - \frac{2}{f} - \frac{2}{R} \Rightarrow \frac{1}{10} = \frac{2}{60} + \frac{2}{R} \Rightarrow R = 30 c m$

Now using $\frac{1}{f} = (μ - 1) (\frac{1}{R})$ , we get $μ = 1.5$

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A ray of light travels from an optically denser to rarer medium. The critical angle for the two media is C. The maximum possible deviation of the ray will be

A.

$(\frac{π}{2} - C)$

B.

2C

C.

$π - 2 C$

D.

$π - C$

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Explanation

When light travels from an optically denser to rarer medium, the maximum possible deviation occurs when the angle of incidence is equal to the critical angle. The refracted ray grazes the surface, and the angle between the incident and refracted rays is (Ï€ - 2C), where C is the critical angle.

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An astronaut is looking down on earth's surface from a space shuttle at an altitude of $400 k m$ . Assuming that the astronaut's pupil diameter is 5 mm and the wavelength of visible light is 500 nm. The astronaut will be able to resolve linear object of the size of about

A.

0.5 m

B.

5 m

C.

50 m

D.

500 m

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Explanation

(c) $\frac{x}{r} = \frac{1.22 λ}{d} \Rightarrow x = \frac{1.22 λ r}{d} = \frac{1.22 \times 500 \times 10^{- 9} \times 400 \times 10^{3}}{5 \times 10^{- 3}} = 50 m$

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The average distance between the earth and moon is $38.6 \times 10^{4}$ km. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of 5 m with $λ = 6000 \overset{\circ}{Α}$ is

A.

5.65 m

B.

28.25 m

C.

11.30 m

D.

56.51 m

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Explanation

(d) Resolving power $= \frac{1.22 λ}{a} = \frac{1.22 \times 6000 \times 10^{- 10}}{5}$
Also resolving power $= \frac{d}{D} = \frac{d}{38.6 \times 10^{7}}$

$∴ \frac{1.22 \times 6 \times 10^{- 7}}{5} = \frac{d}{38.6 \times 10^{7}} \Rightarrow d = \frac{1.22 \times 6 \times 10^{- 7} \times 38.6 \times 10^{7}}{5} m = 56.51 m$

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The distance of the moon from earth is $3.8 \times 10^{5} k m$ . The eye is most sensitive to light of wavelength 5500 Å. The minimum separation between two points on the moon that can be resolved by a 500 cm telescope will be

A.

51 m

B.

60 m

C.

70 m

D.

All the above

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Explanation

(a) As limit of resolution
$∆ θ = \frac{1}{R e s o l v i n g P o w e r (R P)}$ ;

and if x is the distance between points on the surface of moon which is at a distance r from the telescope.

$∆ θ = \frac{x}{r}$
$∆ θ = \frac{1}{R P} = \frac{x}{r} x = \frac{r}{R P} = \frac{r}{d / 1.22 λ} \Rightarrow x = \frac{1.22 λ r}{d}$

$= \frac{1.22 \times 5500 \times 10^{- 10} \times (3.8 \times 10^{8})}{500 \times 10^{- 2}} = 51 m$

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A point object is moving on the principal axis of a concave mirror of focal length 24 cm towards the mirror. When it is at a distance of 60 cm from the mirror, its velocity is 9cm/sec. What is the velocity of the image at that instant

A.

5cm/sec towards the mirror

B.

4cm/sec towards the mirror

C.

4cm/sec away from the mirror

D.

9cm/sec away from the mirror

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Explanation

(c) $v_{i} = - {(\frac{f}{f - u})}^{2} v_{0} = - {(\frac{- 24}{- 24 - (- 60)})}^{2} \times 9 = 4 c m / s e c$ .

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A concave mirror is placed on a horizontal table with its axis directed vertically upwards. Let O be the pole of the mirror and C is its centre of curvature. A point object is placed at C. It has a real image, also located at C. If the mirror is now filled with water, the image will be :

A.

Real, and will remain at C

B.

Real, and located at a point between C and $\infty$

C.

Virtual and located at a point between C and O

D.

Real, and located at a point between C and O

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We wish to see inside an atom. Assuming the atom to have a diameter of 100 pm, this means that one must be able to resolved a width of say 10 p.m. If an electron microscope is used, the minimum electron energy required is about

A.

1.5 KeV

B.

15 KeV

C.

150 KeV

D.

1.5 KeV

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Explanation

(b) Wave length of the electron wave be $10 \times 10^{- 12} m$ ,
Using $λ = \frac{h}{\sqrt{2 m E}} \Rightarrow E = \frac{h^{2}}{λ^{2} \times 2 m}$
$= \frac{{(6.63 \times 10^{- 34})}^{2}}{{(10 \times 10^{- 12})}^{2} \times 2 \times 9.1 \times 10^{- 31}} J o u l e$

$= \frac{{(6.63 \times 10^{- 34})}^{2}}{{(10 \times 10^{- 12})}^{2} \times 2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19}} e V = 15.1 K e V$

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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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