Wave Optics — Page 5 | JEE Physics

Q41

In a Young's double slits experiment, the ratio of amplitude of light coming from slits is

2: 1

. The ratio of the maximum to minimum intensity in the interference pattern is:

A 25 : 9

B 9 : 1

C 9 : 4

D 2 : 1

Correct Answer

Option B

Solution

Given the amplitude ratio,

\frac{A_1}{A_2} = \frac{2}{1}

, we can find the maximum and minimum intensities using the formula:

\frac{I_{max}}{I_{min}} = \left(\frac{A_1 + A_2}{A_1 - A_2}\right)^2

Substituting the given amplitude ratio:

\frac{I_{max}}{I_{min}} = \left(\frac{2 + 1}{2 - 1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9

Thus, the ratio of maximum to minimum intensity in the interference pattern is:

\frac{I_{max}}{I_{min}} = 9:1

Q42

The width of fringe is

2 \mathrm{~mm}

on the screen in a double slits experiment for the light of wavelength of

400 \mathrm{~nm}

. The width of the fringe for the light of wavelength 600

\mathrm{nm}

will be:

A 4 mm

B 1.33 mm

C 2 mm

D 3 mm

Correct Answer

Option D

Solution

In the double-slit experiment, the fringe width ( $\beta$ ) is given by the formula : $\beta = \dfrac{\lambda D}{d}$ where: $\lambda$ is the wavelength of the light, $D$ is the distance between the screen and the double-slit, $d$ is the separation between the two slits.

If we are considering a change in wavelength but the fringe width is changing and the setup of the experiment (the values of $D$ and $d$ ) stays the same, we can see that the fringe width is directly proportional to the wavelength.

This is because $D$ and $d$ are constants in this case, so we can write : $\dfrac{\beta_1}{\beta_2} = \dfrac{\lambda_1}{\lambda_2}$ Here, the given wavelengths are $\lambda_1 = 400 \, \text{nm}$ and $\lambda_2 = 600 \, \text{nm}$ , and the given fringe width for the light of wavelength $400 \, \text{nm}$ is $\beta_1 = 2 \, \text{mm}$ .

We are asked to find the fringe width $\beta_2$ for the light of wavelength $600 \, \text{nm}$ .

Substituting the given values into the proportionality equation, we get : $\dfrac{2 \, \text{mm}}{\beta_2} = \dfrac{400 \, \text{nm}}{600 \, \text{nm}}$ Solving this equation for $\beta_2$ gives: $\beta_2 = 2 \, \text{mm} \times \dfrac{600 \, \text{nm}}{400 \, \text{nm}} = 3 \, \text{mm}$

Q43

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is

7 \lambda / 4

. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is :

A

\dfrac{1}{2}

B

\dfrac{3}{4}

C

\dfrac{1}{3}

D

\dfrac{1}{4}

Correct Answer

Option A

Solution

\begin{aligned} & \Delta x=\frac{7 \lambda}{4} \\ & \phi=\frac{2 \pi}{\lambda} \Delta \mathrm{x}=\frac{2 \pi}{\lambda} \times \frac{7 \lambda}{4}=\frac{7 \pi}{2} \\ & \mathrm{I}=\mathrm{I}_{\max } \cos ^2\left(\frac{\phi}{2}\right) \\ & \frac{\mathrm{I}}{\mathrm{I}_{\max }}=\cos ^2\left(\frac{\phi}{2}\right)=\cos ^2\left(\frac{7 \pi}{2 \times 2}\right)=\cos ^2\left(\frac{7 \pi}{4}\right) \\ & =\cos ^2\left(2 \pi-\frac{\pi}{4}\right) \\ & =\cos ^2 \frac{\pi}{4} \\ & =\frac{1}{2} \\ & \end{aligned}

Q44

A beam of unpolarised light of intensity

I_0

is passed through a polaroid

A

and then through another polaroid

B

which is oriented so that its principal plane makes an angle of

45^{\circ}

relative to that of

A

. The intensity of emergent light is:

A

I_0 / 2

B

I_0 / 8

C

I_0 / 4

D

I_0

Correct Answer

Option C

Solution

Intensity of emergent light

=\frac{\mathrm{I}_0}{2} \cos ^2 45^{\circ}=\frac{\mathrm{I}_0}{4}

Q45

Given below are two statements : Statement I : When the white light passed through a prism, the red light bends lesser than yellow and violet. Statement II : The refractive indices are different for different wavelengths in dispersive medium. In the light of the above statements, chose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Statement I is true but Statement II is false

C Statement I is false but Statement II is true

D Both Statement I and Statement II are false

Correct Answer

Option A

Solution

The correct answer is Option A: Both Statement I and Statement II are true.

Explanation: Statement I discusses the dispersion of white light through a prism, which is a phenomenon where white light splits into its component colors when passed through a prism.

This happens because different colors (or wavelengths) of light bend (refract) by different amounts upon passing through a prism.

Red light bends the least while violet bends the most, with yellow falling somewhere in between.

This is why the statement, "When the white light passed through a prism, the red light bends lesser than yellow and violet," is true.

Statement II addresses the underlying cause of the dispersion of light, which is that the refractive index of a medium varies with wavelength (or color) of light.

This property of the medium is known as dispersion.

A refractive index determines how much light bends when entering a medium.

Since the refractive index varies with the wavelength, different colors of light bend by different amounts when they pass through a dispersive medium like a glass prism.

This is why violet light bends more than red light - because the refractive index for violet light is higher than that for red light in glass.

Therefore, the statement, "The refractive indices are different for different wavelengths in dispersive medium," is also true.

Thus, both statements I and II accurately describe the principles behind the phenomenon of light dispersion through a prism, making Option A the correct choice.

Q46

Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion-(A) : If Young's double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. Reason-(R) : The speed of light reduces in an optically denser medium than air while its frequency does not change. In the light of the above statements, choose the most appropriate answer from the options given below :

A (A) is false but (R) is true

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C Both (A) and (R) are true and (R) is the correct explanation of (A)

D (A) is true but (R) is false

Correct Answer

Option C

Solution

We know, in YDSE, $\beta$ (fringe width)

= {{\lambda D} \over d}

In denser medium,

\lambda \downarrow \Rightarrow \beta \downarrow

$\Rightarrow$ fringes come closer. Also,

\mu = {c \over v} \Rightarrow v = {c \over \mu }

\Rightarrow \mu \uparrow \Rightarrow v \downarrow \Rightarrow

speed of light reduces. frequency remains same,

\Rightarrow \mu = {{{\lambda _{vac.}}f} \over {{\lambda _{med.}}f}} \Rightarrow {\lambda _{med}} = {{{\lambda _{vac.}}} \over \mu }

(as

v = \lambda f

) Hence, option C is correct.

Q47

A transparent film of refractive index, 2.0 is coated on a glass slab of refractive index, 1.45. What is the minimum thickness of transparent film to be coated for the maximum transmission of Green light of wavelength 550 nm . [Assume that the light is incident nearly perpendicular to the glass surface.]

A 68.7 nm

B 137.5 nm

C 94.8 nm

D 275 nm

Correct Answer

Option B

Solution

For transmitted green light to be maxima, reflected green should be minima.

\begin{aligned} & \Delta \mathrm{P}=2 \mu_0 \mathrm{t}=\mathrm{n} \lambda \\ & \Rightarrow \mathrm{t}=\frac{\mathrm{n} \lambda}{2 \mu_0} \therefore \mathrm{t}_{\min }=\frac{\lambda}{2 \mu_0}=\frac{550}{2 \times 2}=137.5 \end{aligned}

Q48

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : In Young's double slit experiment, the fringes produced by red light are closer as compared to those produced by blue light. Reason (R) : The fringe width is directly proportional to the wavelength of light. In the light of the above statements, choose the correct answer from the options given below :

A (A) is true but (R) is false

B Both (A) and (R) are true and (R) is the correct explanation of (A)

C (A) is false but (R) is true

D Both

(\mathbf{A})

and

(\mathbf{R})

are true but

(\mathbf{R})

is NOT the correct explanation of (A)

Correct Answer

Option C

Solution

In Young's double-slit experiment, the width of the fringes, denoted as $\beta$ , can be calculated using the formula: $\beta = \dfrac{\lambda D}{d}$ In this formula: $\lambda$ represents the wavelength of the light used. $D$ is the distance from the slits to the screen. $d$ is the distance between the slits.

According to this relationship, the fringe width $\beta$ is directly proportional to the wavelength $\lambda$ .

This means that as the wavelength increases, the fringe width also increases.

When comparing red and blue light: Red light has a longer wavelength ( $\lambda_R$ ) compared to blue light ( $\lambda_b$ ).

Therefore, the fringe width $\beta_R$ for red light is greater than the fringe width $\beta_b$ for blue light, as $\lambda_R > \lambda_b$ .

This shows that with red light, the fringes are actually farther apart compared to the fringes produced by blue light.

Thus, the assertion that red light fringes are closer than those produced by blue light is false, while the reason about fringe width being proportional to the wavelength is true.

Q49

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is

x \mathrm{~d}

. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is

9: 4

then what is the value of

x

? (Assume that the field strength varies according to the slit width.)

A 2

B 3

C 4

D 5

Correct Answer

Option D

Solution

Let the amplitude from the slit of width

d

be proportional to

E

and from the slit of width

xd

be proportional to

xE

(with

x>1

, as is evident from the given options). For two coherent waves with amplitudes

E_1

and

E_2

, the resultant intensity when added in phase (maximum) and out of phase (minimum) is given by

I_{\max} \propto (E_1+E_2)^2 \quad \text{and} \quad I_{\min} \propto (E_2-E_1)^2.

Here, setting

E_1 = E

(from the narrower slit of width

d

) and

E_2 = xE

(from the wider slit of width

xd

), we have

I_{\max} \propto (E+xE)^2 = E^2 (1+x)^2,

and since

x > 1

the subtraction in the destructive case gives

I_{\min} \propto (xE-E)^2 = E^2 (x-1)^2.

The ratio of maximum to minimum intensity is therefore

\frac{I_{\max}}{I_{\min}} = \frac{(1+x)^2}{(x-1)^2}.

We are given that

\frac{(1+x)^2}{(x-1)^2} = \frac{9}{4}.

Taking the square root of both sides (noting that all quantities are positive) leads to

\frac{1+x}{x-1} = \frac{3}{2}.

To solve for

x

, cross-multiply:

2(1+x) = 3(x-1).

Expanding both sides:

2 + 2x = 3x - 3.

Rearrange the equation to isolate

x

:

2 + 2x + 3 = 3x \quad \Longrightarrow \quad 5 + 2x = 3x,

which implies

x = 5.

Thus, the value of

x

is

\boxed{5}.

Q50

Young's double slit inteference apparatus is immersed in a liquid of refractive index 1.44. It has slit separation of 1.5 mm . The slits are illuminated by a parallel beam of light whose wavelength in air is 690 nm . The fringe-width on a screen placed behind the plane of slits at a distance of 0.72 m , will be:

A 0.63 mm

B 0.33 mm

C 0.46 mm

D 0.23 mm

Correct Answer

Option D

Solution

In Young's double-slit experiment, the fringe width ( $\beta$ ) can be calculated using the formula:

\beta = \frac{\lambda' \cdot D}{d}

where: $\lambda'$ is the wavelength of light in the medium, $D$ is the distance between the slits and the screen, $d$ is the separation between the slits.

The wavelength in the medium ( $\lambda'$ ) is given by:

\lambda' = \frac{\lambda_0}{n}

where: $\lambda_0 = 690 \, \text{nm}$ is the wavelength of light in air, $n = 1.44$ is the refractive index of the liquid.

Substituting the values,

\lambda' = \frac{690 \, \text{nm}}{1.44} \approx 479.17 \, \text{nm}

Now, convert $\lambda'$ to meters:

\lambda' = 479.17 \, \text{nm} = 479.17 \times 10^{-9} \, \text{m}

Next, use the values for $D$ and $d$ : $D = 0.72 \, \text{m}$ $d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$ Calculating the fringe width:

\beta = \frac{479.17 \times 10^{-9} \, \text{m} \cdot 0.72 \, \text{m}}{1.5 \times 10^{-3} \, \text{m}}

\beta \approx \frac{344.2 \times 10^{-9} \, \text{m}}{1.5 \times 10^{-3} \, \text{m}}

\beta \approx 0.229 \times 10^{-3} \, \text{m} = 0.229 \, \text{mm}

The closest answer choice is: Option D: 0.23 mm