Wave Optics — Page 7 | JEE Physics

Q61

The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1

\mu

m. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

A 100

\mu

m

B 25

\mu

m

C 50

\mu

m

D 75

\mu

m

Correct Answer

Option B

Solution

Given 2 $\theta$ = 60o $\Rightarrow$ $\theta$ = 30o We know,

a

sin $\theta$ = n $\lambda$ for first minima n = 1, $\therefore$ At first minima

a

sin = $\lambda$ $\Rightarrow$

\,\,\,

10 $-$ 6 $\times$ sin 30o = $\lambda$ $\Rightarrow$

\,\,\,

$\lambda$ =

{{{{10}^{ - 6}}} \over 2}

m Now after making a new slit, $\therefore$ Fringe width, $\beta$ =

{{\lambda D} \over d}

given, D = 50 cm and $\beta$ = 1 cm. $\therefore$ 1 $\times$ 10 $-$ 2 =

{{0.5 \times {{10}^{ - 6}} \times 50 \times {{10}^{ - 2}}} \over d}

$\Rightarrow$ d = 25 $\times$ 10 $-$ 6 m = 25 $\mu$ m.

Q62

In a Young's double slit experiment, the source is white light. One of the slits is covered by red filter and another by a green filter. In this case:

A there shall be alternate interference fringes of red and green.

B there shall be an interference pattern for red distinct from that for green.

C there shall be an interference pattern, where each fringe's pattern center is green and outer edges is red.

D there shall be no interference fringes.

Correct Answer

Option D

Solution

When white light is used in Young's double slit experiment and one slit is covered by a red filter and the other by a green filter, several effects occur: Different Fringe Widths: Each color will produce fringes of different widths due to their varying wavelengths.

This means that red and green will not overlap perfectly.

Color Overlap: As the fringe patterns develop, some regions will have overlap due to the differing fringe spacings of red and green light.

This overlap will create areas where the colors mix, potentially leading to an appearance of intermediate colors.

Overall, because each color produces its own pattern and these do not align, a clear interference pattern as seen with monochromatic light (a single color) will not be formed.

Instead, there will be a complex pattern where the fringes do not distinctly appear as solid bands of red or green due to the overlapping and different fringe widths.

Q63

A light whose electric field vectors are completely removed by using a good polaroid, allowed to incident on the surface of the prism at Brewster's angle. Choose the most suitable option for the phenomenon related to the prism.

A Reflected and refracted rays will be perpendicular to each other.

B Wave will propagate along the surface of prism.

C No refraction, and there will be total reflection of light.

D No reflection, and there will be total transmission of light.

Correct Answer

Option D

Solution

When electric field vector is completely removed and incident on Brewster's angle then only refraction takes place.

Q64

A light wave is propagating with plane wave fronts of the type

x+y+z=

constant. Th angle made by the direction of wave propagation with the

x

-axis is :

A

\cos ^{-1}(2 / 3)

B

\cos ^{-1}\left(\dfrac{1}{\sqrt{3}}\right)

C

\cos ^{-1}\left(\dfrac{1}{3}\right)

D

\cos ^{-1}\left(\sqrt{\dfrac{2}{3}}\right)

Correct Answer

Option B

Solution

The light wave is propagating with plane wave fronts described by the equation $x + y + z = \text{constant}$ .

The wave propagates in a direction perpendicular to these wave fronts.

This direction is symmetric with respect to the $x$ , $y$ , and $z$ axes.

Since the wave is symmetric about these axes, the angle it makes with each axis is the same.

Let these angles be $\alpha$ , $\beta$ , and $\gamma$ , which are the angles made by the light with the $x$ , $y$ , and $z$ axes, respectively.

This implies: $\cos \alpha = \cos \beta = \cos \gamma$ According to the sum of the squares of the direction cosines: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ Substituting the equality of direction cosines, we get: $3 \cos^2 \alpha = 1$ $\cos^2 \alpha = \dfrac{1}{3}$ Thus, the angle $\alpha$ is given by: $\alpha = \cos^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$

Q65

In Young's double slits experiment, the position of 5

\mathrm{^{th}}

bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is :

A 60

\mu

m

B 48

\mu

m

C 36

\mu

m

D 12

\mu

m

Correct Answer

Option A

Solution

In Young's double-slit experiment, the distance between the slits and the screen, $L = 1 \text{ m}$ , the wavelength of the light, $\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$ , and the position of the 5th bright fringe from the central maximum, $y = 5 \text{ cm} = 0.05 \text{ m}$ .

The distance between the central maximum and the $n$ th bright fringe is given by the formula:

y_n = \frac{n\lambda L}{d}

where $d$ is the distance between the two slits. We can rearrange this formula to solve for $d$ :

d = \frac{n\lambda L}{y_n}

Substituting the values given in the question, we get:

d = \frac{5 \times 600 \times 10^{-9} \times 1 \times 100}{5}

d = 6 \times 10^{-5} \text{ m}

d = 60 \ \mu\text{m}

Therefore, the separation between the slits is $60 \ \mu\text{m}$ .

Q66

The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000

\mathop A\limits^o

is used, the minimum separation between two points, to be seen as distinct, will be :

A 0.12

\mu

m

B 0.38

\mu

m

C 0.24

\mu

m

D 0.48

\mu

m

Correct Answer

Option C

Solution

Numerical aperature of the microscope is given as

NA = {{0.61\lambda } \over d}

Where d = minimum sparaton between two points to be seen as distinct

d = {{0.61\lambda } \over {NA}} = {{\left( {0.61} \right) \times \left( {5000 \times 10\,{m^{ - 10}}} \right)} \over {1.25}}

= 2.4

\times {10^{ - 7}}\,m = 0.24\,\mu m

Q67

A microwave of wavelength

2.0 \mathrm{~cm}

falls normally on a slit of width

4.0 \mathrm{~cm}

. The angular spread of the central maxima of the diffraction pattern obtained on a screen

1.5 \mathrm{~m}

away from the slit, will be :

A

60^{\circ}

B

45^{\circ}

C

15^{\circ}

D

30^{\circ}

Correct Answer

Option A

Solution

To determine the angular spread of the central maximum in a single-slit diffraction pattern, we can use the formula for the angular position of the first minimum (also understood as the boundary of the central maximum) on either side of the center.

For a single-slit diffraction pattern, the angle $\theta$ to the first minimum is given by the condition:

a \sin(\theta) = m\lambda

where :

a

is the width of the slit, $\lambda$ is the wavelength of the light (or microwaves in this case),

m

is the order number of the minimum with

m = \pm1, \pm2, \pm3, ...

(for the first minimum,

m = \pm1

) However, we are only interested in the angle to the first minimum, so we will only consider

m = \pm1

. Since the slit width

a = 4.0 \mathrm{cm}

and the wavelength

\lambda = 2.0 \mathrm{cm}

, we substitute these values into the equation to find $\theta$ :

4.0 \mathrm{cm} \times \sin(\theta) = 1 \times 2.0 \mathrm{cm}

$\Rightarrow$

4.0 \sin(\theta) = 2.0

$\Rightarrow$

\sin(\theta) = \frac{2.0}{4.0}

$\Rightarrow$

\sin(\theta) = 0.5

The angle whose sine is 0.5 is

30^{\circ}

. This angle of

30^{\circ}

is the angle from the center to the first minimum on one side.

The angular spread of the central maximum would cover the range from the first minimum on one side to the first minimum on the other side, totalling twice this angle:

\text{Angular spread} = 2 \times \theta = 2 \times 30^{\circ} = 60^{\circ}

Therefore, the angular spread of the central maxima of the diffraction pattern is

60^{\circ}

, which corresponds to Option A.

Q68

Light of wavelength

550

nm

falls normally on a slit of width

22.0 \times {10^{ - 5}}

cm.

The angular position of the second minima from the central maximum will (in radians) :

A

{\pi \over {12}}

B

{\pi \over 8}

C

{\pi \over 6}

D

{\pi \over 4}

Correct Answer

Option C

Solution

Angular position of nth minima from central maxima, sin $\theta$ =

{{n\lambda } \over a}

here n = 2

\therefore\,\,\,\,

sin $\theta$ =

{{2 \times 550 \times {{10}^{ - 9}}} \over {22 \times {{10}^{ - 5}}}}

=

{1 \over 2}

\therefore\,\,\,

$\theta$ =

{\pi \over 6}

rad.

Q69

The ratio of intensities at two points

\mathrm{P}

and

\mathrm{Q}

on the screen in a Young's double slit experiment where phase difference between two waves of same amplitude are

\pi / 3

and

\pi / 2

, respectively are

A 2 : 3

B 1 : 3

C 3 : 1

D 3 : 2

Correct Answer

Option D

Solution

In a Young's double slit experiment, the intensity at a point on the screen is given by the formula :

I = 4I_0\cos^2\left(\frac{\pi d\sin\theta}{\lambda}\right)

where $I_0$ is the intensity at the center of the pattern, $d$ is the distance between the slits, $\theta$ is the angle between the line joining the point to the center of the pattern and the line passing through the center of the pattern and the slits, and $\lambda$ is the wavelength of the light used.

The phase difference between the waves from the two slits at a point on the screen is given by :

\Delta \phi = \frac{2\pi d\sin\theta}{\lambda}

For a phase difference of $\pi/3$ between the waves from the two slits at point P, we have :

\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{3}

For a phase difference of $\pi/2$ between the waves from the two slits at point Q, we have :

\frac{\Delta \phi}{\pi} = \frac{2d\sin\theta}{\lambda} = \frac{1}{2}

Solving for $\sin\theta$ in both cases, we get:

\sin\theta_P = \frac{\lambda}{6d},\quad \sin\theta_Q = \frac{\lambda}{4d}

Substituting these values in the formula for intensity, we get :

\frac{I_P}{I_Q} = \frac{4\cos^2\left(\frac{\pi}{6}\right)}{4\cos^2\left(\frac{\pi}{4}\right)} = \frac{3}{2}

Therefore, the ratio of intensities at points P and Q is 3 : 2

Q70

A mixture of light, consisting of wavelength

590

nm

and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the

4

th bright fringe of the unknown light. From this data, the wavelength of the unknown light is :

A

885.0

nm

B

442.5

nm

C

776.8

nm

D

393.4

nm

Correct Answer

Option B

Solution

Third bright fringe of known light coincides with the 4th bright fringe of the unknown light. $\therefore$

{{3\left( {590} \right)D} \over d} = {{4\lambda D} \over d}

\Rightarrow \lambda = {3 \over 4} \times 590

= 442.5\,nm