Geometrical Optics — Page 6 | NEET Physics

Q51

A ray of light is incident at an angle of incidence i, on one face of a prism of angle A (assumed to be small ) and emerges normally from the opposite face. If the refractive index of the prism is

\mu

. the angle of incidence i, is nearly equal to

A

\mu

A

B

{{\mu A} \over 2}

C

{A \over \mu }

D

{A \over {2\mu }}

Correct Answer

Option A

Solution

For normally emerge e = 0 Therefore r 2 = 0 and r 1 = A Snell’s Law for Incident ray’s 1sin i = $\mu$ sin r 1 = $\mu$ sin A For small angle i = $\mu$ A

Q52

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

A equal to that of glass

B less than one

C greater than that of glass

D less than that of glass

Correct Answer

Option A

Solution

According to lens maker's formula

{1 \over f} = \left( {{{{\mu _g}} \over {{\mu _L}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)

As the biconvex lens dipped in a liquid acts as a plane sheet of glass, therefore f = $\infty$ $\Rightarrow$

{1 \over f} = 0

$\therefore$

{{{{\mu _g}} \over {{\mu _L}}} - 1 = 0}

$\Rightarrow$

{{\mu _g} = {\mu _L}}

Q53

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 equal

A

{f_1} + {f_2}

B

-

{f_1} + {f_2}

C 2

{f_1} + {f_2}

D

-

2

{f_1} + {f_2}

Correct Answer

Option C

Solution

$\therefore$ d = 2f 1 + f 2

Q54

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

A 10 cm, 10 cm

B 15 cm, 5 cm

C 18 cm, 2 cm

D 11 cm, 9 cm

Correct Answer

Option C

Solution

Magnifying power, m =

{{{f_0}} \over {{f_e}}}

= 9 ..........(1) where fo and fe are the focal lengths of the objective and eyepiece respectively Also,f o + f e = 20 cm ...(

2) On solving (1) and (2), we get f o = 18 cm, f e = 2 cm

Q55

A converging beam of rays is incident on a diverging lens. Having passed through the lens the rays intersect at a point 15 cm from the lens on the opposite side. If the lens is removed the point where the rays meet will move 5 cm closer to the lens. The focal length of the lens is

A 5 cm

B

-

10 cm

C 20 cm

D

-

30 cm

Correct Answer

Option D

Solution

Here, v = +15 cm, u = +(15 – 5) = +10 cm According to lens formula

{1 \over v} - {1 \over u} = {1 \over f}

$\Rightarrow$

{1 \over {15}} - {1 \over {10}} = {1 \over f}

$\Rightarrow$ f = -30 cm

Q56

A thin prism of angle 15 o made of glass of refractive index

\mu

1 = 1.5 is combined with another prism of glass of refractive index

\mu

2 = 1.75. The combination of the prisms produces dispersion without deviation. The angle of the second prism should be

A 5 o

B 7 o

C 10 o

D 12 o

Correct Answer

Option C

Solution

For without deviation

{A \over {A'}} = {{\mu ' - 1} \over {\mu - 1}}

$\Rightarrow$

{{15} \over {A'}} = {{1.75 - 1} \over {1.50 - 1}}

$\Rightarrow$ A' = 10 o

Q57

Which of the following is not due to total internal reflection ?

A Working of optical fibre

B difference between apparent and real depth of a pond

C Mirage on hot summer days

D Brilliance of diamond

Correct Answer

Option B

Solution

Difference between apparent and real depth of a pond is due to refraction.

Other three are due to total internal reflection.

Q58

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

A Virtual, upright, height = 1 cm

B Virtual, upright, height = 0.5 cm

C Real, inverted, height = 4 cm

D Real, inverted, height = 1 cm

Correct Answer

Option C

Solution

Focal length of the lens

{1 \over f} = \left( {1.5 - 1} \right)\left( {{1 \over {20}} - {1 \over { - 20}}} \right)

$\Rightarrow$ f = 20 cm From lens formula

{1 \over v} - {1 \over u} = {1 \over f}

$\Rightarrow$

{1 \over v} - {1 \over { - 30}} = {1 \over {20}}

$\Rightarrow$ v = 60 cm

{I \over O} = m = {v \over u}

=

{{60} \over { - 30}}

= -2 $\therefore$ I = -2 $\times$ (O) = -2 $\times$ 2 = -4 cm So image will be real inverted and of size 4 cm.

Q59

A rays of light is incident on a 60 o prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is

A zero

B 30 o

C 45 o

D 60 o

Correct Answer

Option B

Solution

Angle of prism, A = r 1 + r 2 For minimum deviation r 1 = r 2 = r $\therefore$ A = 2r Given, A = 60° Hence, r =

{A \over 2}

=

{{60^\circ } \over 2}

= 30 o

Q60

The speed of light in media M 1 and M 2 are 1.5

\times

10 8 m/s and 2.0

\times

10 8 m/s respectively. A ray of light enters from medium M 1 to M 2 at an incidence angle i. If the rays suffers total internal reflection, the value of i is

A Equal to sin

-

1

\left( {{2 \over 3}} \right)

B Equal to or less than sin

-

1

\left( {{3 \over 5}} \right)

C Equal to or greater than sin

-

1

\left( {{3 \over 4}} \right)

D Less than sin

-

1

\left( {{2 \over 3}} \right)

Correct Answer

Option C

Solution

Refractive index for medium M 1 is $\mu$ 1 =

{{3 \times {{10}^8}} \over {1.5 \times {{10}^8}}}

= 2 Refractive index for medium M 2 is $\mu$ 2 =

{{3 \times {{10}^8}} \over {2.0 \times {{10}^8}}}

=

{3 \over 2}

For total internal reflection, sini $\ge$ sinC But sinC =

{{{\mu _2}} \over {{\mu _1}}}

sini $\ge$

{{{\mu _2}} \over {{\mu _1}}}

$\ge$

{{3/2} \over 2}

$\Rightarrow$ i $\ge$ sin $-$ 1

\left( {{3 \over 4}} \right)