Wave Optics — Page 7 | NEET Physics

Q61

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: $\alpha = \cos^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$

Q62

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 $60 \ \mu\text{m}$ .

Q63

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

Q64

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.

Q65

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.

Q66

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

Q67

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

Q68

Two coherent sources of light interfere. The intensity ratio of two sources is

1: 4

. For this interference pattern if the value of

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

is equal to

\dfrac{2 \alpha+1}{\beta+3}

, then

\dfrac{\alpha}{\beta}

will be :

A 1.5

B 2

C 0.5

D 1

Correct Answer

Option B

Solution

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

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

$\therefore$

{{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{2({I_1} + {I_2})} \over {4 \times \sqrt {{I_1}{I_2}} }}

= {1 \over 2} \times {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {\sqrt {{{{I_1}} \over {{I_2}}}} }}

= {1 \over 2} \times {{\left( {{1 \over 4} + 1} \right)} \over {\left( {{1 \over 2}} \right)}}

= {5 \over 4} = {{2 \times 2 + 1} \over {1 + 3}}

$\therefore$

{\alpha \over \beta } = {2 \over 1} = 2

Q69

In a Young's double slit experiment, the separation between the slits is 0.15 mm. in the experiment, a source of light of wavelengh 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is :

A 4.9 mm

B 5.9 mm

C 6.9 mm

D 3.9 mm

Correct Answer

Option B

Solution

\beta = {{\lambda D} \over d} = {{589 \times {{10}^{ - 9}} \times 1.5} \over {0.15 \times {{10}^{ - 3}}}}

= 5.9 mm

Q70

A monochromatic light of wavelength

6000 ~\mathring{A}

is incident on the single slit of width

0.01 \mathrm{~mm}

. If the diffraction pattern is formed at the focus of the convex lens of focal length

20 \mathrm{~cm}

, the linear width of the central maximum is :

A

12 \mathrm{~mm}

B

24 \mathrm{~mm}

C

60 \mathrm{~mm}

D

120 \mathrm{~mm}

Correct Answer

Option B

Solution

To find the linear width of the central maximum in a single-slit diffraction pattern, we can use the formula that relates the position of the first minima on either side of the central maximum.

The angle, $\theta$ , at which the first minimum occurs is given by:

a \sin(\theta) = m\lambda

Where: $a$ is the width of the slit $\lambda$ is the wavelength of monochromatic light $m$ is the order number of the minimum (for the first minimum, $m = \pm 1$ ) In our case, we want to find the position of the first minima to determine the width of the central maximum on the screen.

So we will use $m = 1$ and $m = -1$ which correspond to the first minima on either side of the central peak.

Given the width of the slit $a = 0.01$ mm, which we need to convert to meters for consistency with the wavelength:

a = 0.01 \times 10^{-3} \text{m}

And the wavelength $\lambda = 6000 \mathring{A}$ , also converting to meters:

\lambda = 6000 \times 10^{-10} \text{m}

We're using a convex lens of focal length $f = 20$ cm to project the pattern onto a screen.

We'll need to convert the focal length to meters as well:

f = 20 \times 10^{-2} \text{m}

We can calculate the angle $\theta$ needed for the first minima using the approximation of small angles, where $\sin(\theta) \approx \theta$ :

a \theta = m\lambda

Now, solve for $\theta$ for the first minimum (m = 1):

\theta = \frac{\lambda}{a}

Plugging in the values:

\theta = \frac{6000 \times 10^{-10}}{0.01 \times 10^{-3}}

\theta = \frac{6000 \times 10^{-10}}{10^{-5}}

\theta = 6000 \times 10^{-5}

Now, linear width of the central maximum on the screen (from -m to m, or -1 to +1) will be twice the distance from the center to the first minimum, which can be found using the focal length of the lens $f$ and the angle $\theta$ :

y = 2f\theta

Plugging the focal length and calculated theta:

y = 2 \times 20 \times 10^{-2} \times 6000 \times 10^{-5}

y = 40 \times 10^{-2} \times 6000 \times 10^{-5}

y = 240 \times 10^{-3} \text{m}

y = 24 \times 10^{-2} \text{m}

y = 24 \text{mm}

This means the linear width of the central maximum is 24 mm. Hence, the correct answer is Option B.