Wave Optics — Page 9 | NEET Physics

Q81

Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizer A and C is :

A 60o

B 30o

C 45o

D 0o

Correct Answer

Option C

Solution

As after B intensity of light does not drops, it means both A and B are alligned in single line means their plane of polarization is same.

Let C makes an angle $\theta$ with A then C will make $\theta$ with B also, as both A and B are alligned in a single line.

So, after C intensity is =

{{\rm I} \over 2}

cos2 $\theta$ , and , intensity after B =

{{\rm I} \over 2}

cos2 $\theta$ $\times$ cos2 $\theta$ According to question,

{{\rm I} \over 2}

cos4 $\theta$ =

{{\rm I} \over 8}

$\Rightarrow$ \,\,\, CO4 $\theta$ =

{{\rm I} \over 4}

$\Rightarrow$

\,\,\,

cos $\theta$ =

= {1 \over {\sqrt 2 }}

= cos45o

\therefore\,\,\,

$\theta$ = 45o

Q82

In a Young’s double slit experiment, light of 500 nm is used to produce an interference pattern. When the distance between the slits is 0.05 mm, the angular width (in degree) of the fringes formed on the distance screen is close to

A 0.17o

B 1.7o

C 0.57o

D 0.07o

Correct Answer

Option C

Solution

$\beta$ =

{{\lambda D} \over d}

and $\theta$ =

{\beta \over D}

$\Rightarrow$ $\theta$ =

{\lambda \over d}

=

{{500 \times {{10}^{ - 9}}} \over {0.05 \times {{10}^{ - 3}}}}

= 0.01 rad = 0.57o

Q83

A single slit of width b is illuminated by a coherent monochromatic light of wavelength

\lambda

. If the second and fourthminima in the diffraction pattern at a distance 1 m from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum ? (i.e. distance between first minimum on either side of the central maximum)

A 1.5 cm

B 3.0 cm

C 4.5 cm

D 6.0 cm

Correct Answer

Option B

Solution

For single slit diffraction, sin $\theta$ =

{{n\lambda } \over b}

From central maxima the position of nth minima =

{{n\lambda D} \over b}

Now when, n = 2, then x2 =

{{2\lambda D} \over b}

= 0.03 . . . .(1) n = 4, then x4 =

{{4\lambda D} \over b}

= 0.06 . . . .(2) Performing (2) $-$ (1) we get, x4 $-$ x2 =

{{2\lambda D} \over b}

= 0.03

\therefore\,\,\,

{{\lambda D} \over b}

=

{{0.03} \over 2}

As, width of crntral maximum =

{{2\lambda D} \over b}

= 2 $\times$

{{0.03} \over 2}

= 0.03 m = 3 cm

Q84

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :

A 16 : 9

B 25 : 9

C 4 : 1

D 5 : 3

Correct Answer

Option B

Solution

Given that,

{{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}

We know, Imax $=$

{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)^2}

and Imin

= {\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)^2}

$\therefore$

{{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}

$\Rightarrow$

{{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}

$\Rightarrow$

4\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}}

$\Rightarrow$

3\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}}

$\Rightarrow$

{{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}

$\Rightarrow$

{{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}

Q85

To demonstrate the phenomenon of interference, we require two sources which emit radiation

A of nearly the same frequency

B of the same frequency

C of different wavelengths

D of the same frequency and having a definite phase relationship

Correct Answer

Option D

Solution

To observe the phenomenon of interference, you need two sources that emit radiation of the same frequency and have a definite phase relationship.

This is because interference is a result of the superposition of waves, which requires the waves to be coherent.

Coherence is achieved when the waves have a constant phase difference, which implies that they are of the same frequency and have a definite phase relationship (a phase relationship that does not change with time).

So, the correct answer is : Option D : of the same frequency and having a definite phase relationship.

Q86

Given below are two statements : Statement I : If the Brewster's angle for the light propagating from air to glass is

\mathrm{\theta_B}

, then the Brewster's angle for the light propagating from glass to air is

\dfrac{\pi}{2}-\theta_B

Statement II : The Brewster's angle for the light propagating from glass to air is

{\tan ^{ - 1}}({\mu _\mathrm{g}})

where

\mathrm{\mu_g}

is the refractive index of glass. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are false

B Both Statement I and Statement II are true

C Statement I is false but Statement II is true

D Statement I is true but Statement II is false

Correct Answer

Option D

Solution

Case I : Transmitted is $\perp$ to reflected.

$i+r=90^{\circ}$ Snell's law $\mu_{a} \sin i=\mu_{g} \sin r$ $\tan i=\dfrac{\mu_{g}}{\mu_{a}}$ $i=\tan ^{-1}\left(\dfrac{\mu_{g}}{\mu_{a}}\right)=\theta_{B}$ Case II : $i+r=90^{\circ}$ as transmitted is $\perp$ to reflected.

$\tan i=\dfrac{\mu_{a}}{\mu_{g}} \Rightarrow i=\tan ^{-1} \dfrac{\mu_{a}}{\mu_{g}}=\dfrac{\pi}{2}-\theta_{B}$

Q87

A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000

\mathop A\limits^ \circ

and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :

A 3 mm

B 9 mm

C 4.5 mm

D 1.5 mm

Correct Answer

Option B

Solution

The slit width is a = 0.1 mm = 10 $-$ 4 m.

The wavelength of the light is $\lambda$ = 6000 $\times$ 10 $-$ 10 = 6 $\times$ 10 $-$ 7.

The distance from the slit to diffraction bands is D = 0.5 m.

We calculate the third dark band from the central band as follows:

A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}

\Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9

mm

Q88

Visible light of wavelength 6000

\times

10-8 cm falls normally on a single slit and produces a diffraction pattern. It is found that the second diffraction minimum is at 60o from the central maximum. If the first minimum is produced at

\theta

1, then

\theta

1, is close to :

A 45o

B 30o

C 25o

D 20o

Correct Answer

Option C

Solution

For 2nd minima

\sin \theta = {{2\lambda } \over d }

$\Rightarrow$

\sin 60^\circ = {{2\lambda } \over d}

$\Rightarrow$

{\lambda \over d} = {{\sqrt 3 } \over 4}

For 1st minima

\sin {\theta _1} = {\lambda \over d}

=

{{\sqrt 3 } \over 4}

$\Rightarrow$

{\theta _1} =

25o

Q89

In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright fringe at angle

{1 \over {40}}

by using light of wavelength

\lambda

1. When the light of wavelength

\lambda

2 is used a bright fringe is seen at the same angle in the same set up. Given that

\lambda

1 and

\lambda

2 are in visible range (380 nm to 740 nm), their values are -

A 400 nm, 500 nm

B 625 nm, 500 nm

C 380 nm, 500 nm

D 380 nm, 525 nm

Correct Answer

Option B

Solution

Path difference = d sin $\theta$ $\approx$ d $\theta$ = 0.1 $\times$

{1 \over {40}}

mm = 2500nm or bright fringe, path difference must be integral multiple of $\lambda$ . $\therefore$ 2500 = n $\lambda$ 1 = m $\lambda$ 2 $\therefore$ $\lambda$ 1 = 625 (from n = 4), $\lambda$ 2 = 500 (from m = 5)

Q90

Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600nm. coming from a distant object, the limit of resolution of the telescope is close to :-

A 3.0 × 10–7 rad

B 4.5 × 10–7 rad

C 1.5 × 10–7 rad

D 2.0 × 10–7 rad

Correct Answer

Option A

Solution

Limit of resolution =

{{1.22\lambda } \over d}

=

{{1.22 \times 600 \times {{10}^{ - 9}}} \over {250 \times {{10}^{ - 2}}}}

= 2.9 × 10–7 rad.