Geometrical Optics — Page 7 | NEET Physics

Q61

A ray of light travelling in a transparent medium of refractive index

\mu

, falls on a surface separating the medium from air at an angle of incidence of 45 o . For which of the following value of

\mu

the ray can undergo total internal reflection ?

A

\mu

= 1.33

B

\mu

= 1.40

C

\mu

= 1.50

D

\mu

= 1.25

Correct Answer

Option C

Solution

For total internal reflection, sini $>$ sinC where, i = angle of incidence, C = critical angle But, sinC =

{1 \over \mu }

$\therefore$ sini $>$

{1 \over \mu }

$\Rightarrow$ $\mu$ $>$

{1 \over {\sin i}}

$\therefore$ $\mu$ $>$

{1 \over {\sin 45^\circ }}

$\Rightarrow$ $\mu$ $>$

\sqrt 2

Q62

A lens having focal length f and aperture of diameter d forms an image of intensity I. Aperture of diameter

{d \over 2}

in central region of lens is covered by a black paper. Focal length of lens and intensity of image now will be respectively

A

f\,

and

{I \over 4}

B

{{3f} \over 4}

and

{I \over 2}

C

f

and

{{3I} \over 4}

D

{f \over 2}

and

{I \over 2}

Correct Answer

Option C

Solution

By covering aperture, focal length does not change. But intensity is reduced by

{1 \over 4}

times, as aperture diameter

{d \over 2}

is covered. $\therefore$ I' = I -

{I \over 4}

=

{{3I} \over 4}

$\therefore$ New focal length = f and intensity =

{{3I} \over 4}

Q63

Two thin lenses of focal lengths f 1 and f 2 are in contact and coaxial. The power of the combination is

A

{{{f_1} + {f_2}} \over 2}

B

{{{f_1} + {f_2}} \over {{f_1}{f_2}}}

C

\sqrt {{{{f_1}} \over {{f_2}}}}

D

\sqrt {{{{f_2}} \over {{f_1}}}}

Correct Answer

Option B

Solution

{1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}}

$\therefore$ Power P =

{{{f_1} + {f_2}} \over {{f_1}{f_2}}}

Q64

A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length 10 cm. The diameter of the sun is 1.39

\times

10 9 m and its mean distance from the earth is 1.5

\times

10 11 m. What is the diameter of the sun's image on the paper ?

A 6.5

\times

10

-

5 m

B 12.4

\times

10

-

4 m

C 9.2

\times

10

-

4 m

D 6.5

\times

10

-

4 m

Correct Answer

Option C

Solution

\left| {{v \over u}} \right| =

size of image size of object Size of image =

\left| {{v \over u}} \right|

$\times$ Size of object =

\left( {{{{{10}^{ - 1}}} \over {1.5 \times {{10}^{11}}}}} \right) \times \left( {1.39 \times {{10}^9}} \right)

= 9.2 ×10 –4 m $\therefore$ Diameter of the sun’s image = 9.2 ×10 –4 m

Q65

A small coin is resting on the bottom of a beaker filled with liquid . A ray of light from the coin travels upto the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid ?

A 2.4

\times

10 8 m/s

B 3.0

\times

10 8 m/s

C 1.2

\times

10 8 m/s

D 1.8

\times

10 8 m/s

Correct Answer

Option D

Solution

As

{1 \over \mu } = {{\sin i} \over {\sin 90^\circ }}

$\Rightarrow$ $\mu$ =

{1 \over {\sin i}}

=

{5 \over 3}

Speed, v =

{c \over \mu }

=

{{3 \times {{10}^8}} \over {5/3}}

= 1.8 $\times$ 10 8 m/s

Q66

A convex lens and a concave lens, each having same focal length of 25 cm, are put in contact to form a combination of lenses. The power in diopters of the combination is

A zero

B 25

C 50

D infinite.

Correct Answer

Option A

Solution

Power of combination in dioptres P = P 1 + P 2 =

{{100} \over {{f_1}}}

+

{{100} \over {{f_2}}}

=

{{100} \over {25}} - {{100} \over {25}}

= 0

Q67

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 ?

A 2 cm upward

B 1 cm upward

C 4.5 cm downward

D 1 cm downward

Correct Answer

Option B

Solution

Apparent depth =

{{real\,depth} \over \mu }

=

{3 \over {1.5}}

= 2 cm As image appears to be raised by 1 cm, therefore, microscope must be moved upwards by 1 cm.

Q68

A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass slab. When coming out on the opposite parallel face, the red green ray emerge from

A Two points propagating in two different non parallel directions

B Two points propagating in two different parallel directions.

C One point propagating in two different directions.

D One point propagating in the same firections.

Correct Answer

Option B

Solution

Red and green rays emerge from two points, propagating in two different parallel directions.

Q69

The refractive index of the material of a prism is

\sqrt 2

and its refracting angle is 30 o . One of the refracting surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face with retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is

A 45 o

B 60 o

C 0

D 30 o

Correct Answer

Option A

Solution

Using law of triangle, we get

\angle r

= 30 o $\mu$ =

{{\sin i} \over {\sin r}}

$\Rightarrow$

\sqrt 2 \times \sin 30^\circ

= sini $\Rightarrow$ sini =

{1 \over {\sqrt 2 }}

$\Rightarrow$ i = 45 o

Q70

An equiconvex lens is cut into two halves along (i) XOX' and (ii) YOY' as shown in the figure. Let

f,f',

f''

be the focal lengths of the complete lens, of each half in case (i), and of each half in case (ii), respectively. Choose the correct statement from the following

A

f' = f,f'' = 2f

B

f' = 2f,f'' = f

C

f' = f,f'' = f

D

f' = 2f,f'' = 2f

Correct Answer

Option A

Solution

We know from Lens maker's formula

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

Initially, Here R 1 = R and R 2 = –R by convection. $\therefore$

{1 \over f} = \left( {\mu - 1} \right){2 \over R}

$\Rightarrow$

{1 \over {2f}} = \left( {\mu - 1} \right){1 \over R}

If we cut the lens along XOX' then the two halves of the lens will be having the same radii of curvature and so, focal length f' = f .

But when we cut it along YOY' then, we will have R 1 = R but R 2 = $\infty$ $\therefore$

{1 \over {f''}} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over \infty }} \right)

=

\left( {\mu - 1} \right){1 \over R}

=

{1 \over {2f}}

$\Rightarrow$ f'' = 2f