Wave Optics — Page 2 | NEET Physics

Q11

In a Young's double slit experiment, a student observes 8 fringes in a certain segment of screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is

A 6

B 8

C 9

D 12

Correct Answer

Option D

Solution

\beta = {{\lambda D} \over d}

Let length of segment of screen = l

\Rightarrow l = 8{\beta _1} = {{8{\lambda _1}D} \over d}

..... (1) and

l = n{\beta _2} = {{n{\lambda _2}D} \over d}

.... (2) from (1) and (2)

8{\lambda _1} = n{\lambda _2}

8(600 nm) = n(400 nm) n = 12

Q12

Assume that light of wavelength 600 nm is coming from a star. The limit of resolution of telescope whose objective has a diameter of 2 m is :

A 1.83

\times

10 -7 rad

B 7.32

\times

10 -7 rad

C 6.00

\times

10 -7 rad

D 3.66

\times

10 -7 rad

Correct Answer

Option D

Solution

Limit of Resolution for a Telescope is

\theta = 1.22{\lambda \over d}

Given, $\lambda$ = 600 $\times$ 10 -9 ; d = 2 m By substituting the values, we get $\theta$ = 3.66 $\times$ 10 -7 rad

Q13

In Young's double slit experiment, if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled, then the fringe width becomes :

A half

B four times

C one-fourth

D double

Correct Answer

Option B

Solution

Given that the slit distance is made half and the screen distance made double than the original value, then, Fringe width,

\beta = {{\lambda D} \over d}

Now,

d' = {d \over 2}

and

D' = 2D

So,

\beta ' = {{\lambda (2D)} \over {d/2}} = {{4\lambda D} \over d}

$\beta$ = 4 $\beta$

Q14

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1m away, was found to be 0.2 o . What will be the angular width of the first minima, (

\mu

water = 4/3) if the entire experimental apparatus is immersed in water?

A 0.15 o

B 0.05 o

C 0.1 o

D 0.266 o

Correct Answer

Option A

Solution

Angular fringe width (in air) $\theta$ air =

{\beta \over D}

Angular Fringe width (in water) $\theta$ w =

{\beta \over {\mu D}} = {{{\theta _{air}}} \over \mu }

=

{{0.2^\circ } \over {{4 \over 3}}}

= 0.15 o

Q15

In Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength

\lambda

of the light used is 5896

\mathop A\limits^0

and distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20 o . To increase the fringe angular width to 0.21 o (with same

\lambda

and D) the separation between the slits needs to be changed to

A 1.9 mm

B 1.8 mm

C 2.1 mm

D 1.7 mm

Correct Answer

Option A

Solution

Angular width =

{\lambda \over d}

$\therefore$ 0.20 o =

{\lambda \over 2}

$\Rightarrow$ $\lambda$ = 0.2 $\times$ 2 Again, 0.21 o =

{\lambda \over d}

So using value of λ, we have d =

{{0.20 \times 2} \over {0.21}}

= 1.9 mm

Q16

Unpolarised light is incident from air on a plane surface of a material of refractive index

\mu

. At a particular angle of incidence i, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

A Reflected light is polarised with its electric vector parallel to the plane of incidence

B Reflected light is polarised with its electric vector perpendicular to the plane of incidence

C

i = {\sin ^{ - 1}}\left( {{1 \over \mu }} \right)

D i = tan -1

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

Correct Answer

Option B

Solution

If reflected and refracted light rays are perpendicular, reflected light gets polarised with electric field vector perpendicular to the plane of incidence.

Also, tan i = $\mu$ (i = Brewster's angle)

Q17

Young's double slit experiment is first performed in air and then in a medium other than air. It is found that 8 th bright fringe in the medium lies where 5 th dark fringe lies in air. The refractive index of the medium is nearly

A 1.59

B 1.69

C 1.78

D 1.25

Correct Answer

Option C

Solution

Position of 8 th bright fringe in medium x =

{{8{\lambda _m}D} \over {\mu d}}

Position of 5 th dark fringe in air, x' =

{{\left( {5 - {1 \over 2}} \right){\lambda _{air}}D} \over d}

=

{{4.5{\lambda _{air}}D} \over d}

According to question

{{8{\lambda _m}D} \over {\mu d}}

=

{{4.5{\lambda _{air}}D} \over d}

$\Rightarrow$ $\mu$ = 1.78

Q18

The ratio of resolving powers of an optical microscope for two wavelength

\lambda

1 = 4000

\mathop A\limits^ \circ

and

{\lambda _2}

= 6000

\mathop A\limits^ \circ

is

A 9 : 4

B 3 : 2

C 16 : 81

D 8 : 27

Correct Answer

Option B

Solution

The resolving power of an optical microscope, RP =

{{2\mu \sin \theta } \over \lambda }

$\Rightarrow$ RP $\propto$

{1 \over \lambda }

Now,

{{R{P_1}} \over {R{P_2}}} = {{{\lambda _2}} \over {{\lambda _1}}}

=

{{6000} \over {4000}}

=

{3 \over 2}

Q19

Two polaroids P 1 and P 2 are placed with their axis perpendicular to each other. Unpolarised light

I

0 is incident on P 1 . A third polaroid P 3 is kept in between P 1 and P 2 such that its axis makes an angle 45 o with that of P 1 . The intensity of transmitted light through P 2 is

A

{{{I_0}} \over 4}

B

{{{I_0}} \over 8}

C

{{{I_0}} \over 16}

D

{{{I_0}} \over 2}

Correct Answer

Option B

Solution

Apply Malus' Law, Intensity of light from first polaroid P 1 , I 1 =

{{{I_0}} \over 2}

Intensity of light from second polaroid P 3 , I 2 =

{{{I_0}} \over 2}{\cos ^2}45^\circ

=

{{{I_0}} \over 2} \times {1 \over 2}

Intensity of transmitted light through P 2 , I 3 =

{{{I_0}} \over 4}{\cos ^2}45^\circ

=

{{{I_0}} \over 4} \times {1 \over 2}

=

{{{I_0}} \over 8}

Q20

The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio

{{{I_{max}} - {I_{\min }}} \over {{I_{max}} + {I_{min}}}}

will be

A

{{\sqrt n } \over {n + 1}}

B

{{2\sqrt n } \over {n + 1}}

C

{{\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}

D

{{2\sqrt n } \over {{{\left( {n + 1} \right)}^2}}}

Correct Answer

Option B

Solution

Maximum Intensity is given as I max =

{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2}

Minimum intensity is given as I min =

{\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}

Given

{{{I_1}} \over {{I_2}}}

= n

{{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2}

=

{\left( {{{\sqrt {{{{I_1}} \over {{I_2}}}} + 1} \over {\sqrt {{{{I_1}} \over {{I_2}}}} - 1}}} \right)^2}

=

{\left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)^2}

{{{I_{\max }} - {I_{\min }}} \over {{I_{\max }} + {I_{\min }}}} =

{{{{{I_{\max }}} \over {{I_{\min }}}} - 1} \over {{{{I_{\max }}} \over {{I_{\min }}}} + 1}}

=

{{{{\left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)}^2} - 1} \over {{{\left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)}^2} + 1}}

=

{{{{\left( {\sqrt n + 1} \right)}^2} - {{\left( {\sqrt n - 1} \right)}^2}} \over {{{\left( {\sqrt n + 1} \right)}^2} + {{\left( {\sqrt n - 1} \right)}^2}}}

=

{{4\sqrt n } \over {2\left( {n + 1} \right)}}

=

{{2\sqrt n } \over {n + 1}}