QBManager

WAVE OPTICS

283299 In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is

1

36 : 16

2

7 : 5

3

6 : 1

4

36 : 5

Explanation:

: Given that,
$\frac{I_{max}}{I_{min}} = 36$
We know that, amplitude $A \propto \sqrt{I}$
$∴ \frac{I_{max}}{I_{min}} = \frac{(a + b)^{2}}{(a - b)^{2}}$
$\frac{a + b}{a - b} = \sqrt{36}$
$a + b = 6 a - 6 b$
$a : b = 7 : 5$

TS EAMCET 31.07.2022

WAVE OPTICS

283300 In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.

1 2

2 4

3 8

4 16

Explanation:

: Given that,
$d_{1} = d, d_{2} = \frac{d}{2}, D_{1} = D_{2} = D$
$λ_{1} = λ, λ_{2} = 4 λ$
The position of $n^{th}$ order maxima on the screen is-
$x_{n} = \frac{n λ_{1} D_{1}}{d_{1}} = \frac{n λ D}{d}$
And, $x_{n}^{'} = \frac{n λ_{2} D_{2}}{d_{2}}$
$x_{n}^{'} = \frac{n 4 λ \times D}{d / 2}$
$x_{n}^{'} = 8 (\frac{n λ D}{d})$ So, $x_{n}^{'} = 8 x_{n}$

TS EAMCET 20.07.2022

WAVE OPTICS

283302 In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?

1

1 / 9

2

1 / 3

3

1 / 16

4

1 / 2

Explanation:

: Given that,
$I_{min} = I_{0}$
$I_{max} = 4 I_{0}$
We know that,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$4 I_{0} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And,
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{0} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Form equation (i) and equation (ii), we get-
$\frac{4 I_{0}}{I_{0}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 2 (\sqrt{I_{1}} - \sqrt{I_{2}})$
$\sqrt{I_{1}} = 3 \sqrt{I_{2}}$ Hence, $\frac{I_{2}}{I_{1}} = \frac{1}{9}$

WB JEE 2022

WAVE OPTICS

283304 In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $λ$ is $I, λ$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{λ}{4}$ will be

1

\frac{I}{4}

2

\frac{I}{2}

3 I

4 zero

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) Δ x$
Case (i) for path difference $λ$
$ϕ = (\frac{2 π}{λ}) x λ = 2 π$
Case (ii) for path difference $λ / 4$
$ϕ = (\frac{2 π}{λ}) \times \frac{λ}{4} = \frac{π}{2}$
Intensity on screen,
$I = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
For $ϕ = 2 π$
Then, $I = 4 I_{0} \cos^{2} (\frac{2 π}{2})$
Therefore, $I = 4 I_{0}$
For $ϕ = (\frac{π}{2})$ then,
$I^{'} = 4 I_{0} \cos^{2} (\frac{π}{4})$
$I^{'} = 4 I_{0} \times \frac{1}{2}$
$I^{'} = \frac{I}{2} (∵ I_{o} = \frac{I}{4})$

WB JEE 2022

WAVE OPTICS

283299 In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is

1

36 : 16

2

7 : 5

3

6 : 1

4

36 : 5

Explanation:

: Given that,
$\frac{I_{max}}{I_{min}} = 36$
We know that, amplitude $A \propto \sqrt{I}$
$∴ \frac{I_{max}}{I_{min}} = \frac{(a + b)^{2}}{(a - b)^{2}}$
$\frac{a + b}{a - b} = \sqrt{36}$
$a + b = 6 a - 6 b$
$a : b = 7 : 5$

TS EAMCET 31.07.2022

WAVE OPTICS

283300 In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.

1 2

2 4

3 8

4 16

Explanation:

: Given that,
$d_{1} = d, d_{2} = \frac{d}{2}, D_{1} = D_{2} = D$
$λ_{1} = λ, λ_{2} = 4 λ$
The position of $n^{th}$ order maxima on the screen is-
$x_{n} = \frac{n λ_{1} D_{1}}{d_{1}} = \frac{n λ D}{d}$
And, $x_{n}^{'} = \frac{n λ_{2} D_{2}}{d_{2}}$
$x_{n}^{'} = \frac{n 4 λ \times D}{d / 2}$
$x_{n}^{'} = 8 (\frac{n λ D}{d})$ So, $x_{n}^{'} = 8 x_{n}$

TS EAMCET 20.07.2022

WAVE OPTICS

283302 In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?

1

1 / 9

2

1 / 3

3

1 / 16

4

1 / 2

Explanation:

: Given that,
$I_{min} = I_{0}$
$I_{max} = 4 I_{0}$
We know that,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$4 I_{0} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And,
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{0} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Form equation (i) and equation (ii), we get-
$\frac{4 I_{0}}{I_{0}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 2 (\sqrt{I_{1}} - \sqrt{I_{2}})$
$\sqrt{I_{1}} = 3 \sqrt{I_{2}}$ Hence, $\frac{I_{2}}{I_{1}} = \frac{1}{9}$

WB JEE 2022

WAVE OPTICS

283304 In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $λ$ is $I, λ$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{λ}{4}$ will be

1

\frac{I}{4}

2

\frac{I}{2}

3 I

4 zero

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) Δ x$
Case (i) for path difference $λ$
$ϕ = (\frac{2 π}{λ}) x λ = 2 π$
Case (ii) for path difference $λ / 4$
$ϕ = (\frac{2 π}{λ}) \times \frac{λ}{4} = \frac{π}{2}$
Intensity on screen,
$I = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
For $ϕ = 2 π$
Then, $I = 4 I_{0} \cos^{2} (\frac{2 π}{2})$
Therefore, $I = 4 I_{0}$
For $ϕ = (\frac{π}{2})$ then,
$I^{'} = 4 I_{0} \cos^{2} (\frac{π}{4})$
$I^{'} = 4 I_{0} \times \frac{1}{2}$
$I^{'} = \frac{I}{2} (∵ I_{o} = \frac{I}{4})$

WB JEE 2022

WAVE OPTICS

283299 In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is

1

36 : 16

2

7 : 5

3

6 : 1

4

36 : 5

Explanation:

: Given that,
$\frac{I_{max}}{I_{min}} = 36$
We know that, amplitude $A \propto \sqrt{I}$
$∴ \frac{I_{max}}{I_{min}} = \frac{(a + b)^{2}}{(a - b)^{2}}$
$\frac{a + b}{a - b} = \sqrt{36}$
$a + b = 6 a - 6 b$
$a : b = 7 : 5$

TS EAMCET 31.07.2022

WAVE OPTICS

283300 In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.

1 2

2 4

3 8

4 16

Explanation:

: Given that,
$d_{1} = d, d_{2} = \frac{d}{2}, D_{1} = D_{2} = D$
$λ_{1} = λ, λ_{2} = 4 λ$
The position of $n^{th}$ order maxima on the screen is-
$x_{n} = \frac{n λ_{1} D_{1}}{d_{1}} = \frac{n λ D}{d}$
And, $x_{n}^{'} = \frac{n λ_{2} D_{2}}{d_{2}}$
$x_{n}^{'} = \frac{n 4 λ \times D}{d / 2}$
$x_{n}^{'} = 8 (\frac{n λ D}{d})$ So, $x_{n}^{'} = 8 x_{n}$

TS EAMCET 20.07.2022

WAVE OPTICS

283302 In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?

1

1 / 9

2

1 / 3

3

1 / 16

4

1 / 2

Explanation:

: Given that,
$I_{min} = I_{0}$
$I_{max} = 4 I_{0}$
We know that,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$4 I_{0} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And,
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{0} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Form equation (i) and equation (ii), we get-
$\frac{4 I_{0}}{I_{0}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 2 (\sqrt{I_{1}} - \sqrt{I_{2}})$
$\sqrt{I_{1}} = 3 \sqrt{I_{2}}$ Hence, $\frac{I_{2}}{I_{1}} = \frac{1}{9}$

WB JEE 2022

WAVE OPTICS

283304 In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $λ$ is $I, λ$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{λ}{4}$ will be

1

\frac{I}{4}

2

\frac{I}{2}

3 I

4 zero

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) Δ x$
Case (i) for path difference $λ$
$ϕ = (\frac{2 π}{λ}) x λ = 2 π$
Case (ii) for path difference $λ / 4$
$ϕ = (\frac{2 π}{λ}) \times \frac{λ}{4} = \frac{π}{2}$
Intensity on screen,
$I = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
For $ϕ = 2 π$
Then, $I = 4 I_{0} \cos^{2} (\frac{2 π}{2})$
Therefore, $I = 4 I_{0}$
For $ϕ = (\frac{π}{2})$ then,
$I^{'} = 4 I_{0} \cos^{2} (\frac{π}{4})$
$I^{'} = 4 I_{0} \times \frac{1}{2}$
$I^{'} = \frac{I}{2} (∵ I_{o} = \frac{I}{4})$

WB JEE 2022

WAVE OPTICS

283299 In a Young's double slit experiment the intensity ratio between bright and dark fringes is found to be 36. The ratio of amplitude of bright and dark fringes is

1

36 : 16

2

7 : 5

3

6 : 1

4

36 : 5

Explanation:

: Given that,
$\frac{I_{max}}{I_{min}} = 36$
We know that, amplitude $A \propto \sqrt{I}$
$∴ \frac{I_{max}}{I_{min}} = \frac{(a + b)^{2}}{(a - b)^{2}}$
$\frac{a + b}{a - b} = \sqrt{36}$
$a + b = 6 a - 6 b$
$a : b = 7 : 5$

TS EAMCET 31.07.2022

WAVE OPTICS

283300 In a Young's double slit experiment, if the distance between two slits is reduced by a factor of 2 and the wavelength of light is increased 4 times then the distance between two maxima will become times the original value.

1 2

2 4

3 8

4 16

Explanation:

: Given that,
$d_{1} = d, d_{2} = \frac{d}{2}, D_{1} = D_{2} = D$
$λ_{1} = λ, λ_{2} = 4 λ$
The position of $n^{th}$ order maxima on the screen is-
$x_{n} = \frac{n λ_{1} D_{1}}{d_{1}} = \frac{n λ D}{d}$
And, $x_{n}^{'} = \frac{n λ_{2} D_{2}}{d_{2}}$
$x_{n}^{'} = \frac{n 4 λ \times D}{d / 2}$
$x_{n}^{'} = 8 (\frac{n λ D}{d})$ So, $x_{n}^{'} = 8 x_{n}$

TS EAMCET 20.07.2022

WAVE OPTICS

283302 In Young's double slit experiment with a monochromatic light, maximum, intensity is 4 times the minimum intensity in the interference pattern. What is the ratio of the intensities of the two interfering waves?

1

1 / 9

2

1 / 3

3

1 / 16

4

1 / 2

Explanation:

: Given that,
$I_{min} = I_{0}$
$I_{max} = 4 I_{0}$
We know that,
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
$4 I_{0} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2}$
And,
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
$I_{0} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2}$
Form equation (i) and equation (ii), we get-
$\frac{4 I_{0}}{I_{0}} = {(\frac{\sqrt{I_{1}} + \sqrt{I_{2}}}{\sqrt{I_{1}} - \sqrt{I_{2}}})}^{2}$
$\sqrt{I_{1}} + \sqrt{I_{2}} = 2 (\sqrt{I_{1}} - \sqrt{I_{2}})$
$\sqrt{I_{1}} = 3 \sqrt{I_{2}}$ Hence, $\frac{I_{2}}{I_{1}} = \frac{1}{9}$

WB JEE 2022

WAVE OPTICS

283304 In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $λ$ is $I, λ$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{λ}{4}$ will be

1

\frac{I}{4}

2

\frac{I}{2}

3 I

4 zero

Explanation:

: We know that,
Phase difference $(ϕ) = (\frac{2 π}{λ}) \times$ path difference
$= (\frac{2 π}{λ}) Δ x$
Case (i) for path difference $λ$
$ϕ = (\frac{2 π}{λ}) x λ = 2 π$
Case (ii) for path difference $λ / 4$
$ϕ = (\frac{2 π}{λ}) \times \frac{λ}{4} = \frac{π}{2}$
Intensity on screen,
$I = 4 I_{0} \cos^{2} (\frac{ϕ}{2})$
For $ϕ = 2 π$
Then, $I = 4 I_{0} \cos^{2} (\frac{2 π}{2})$
Therefore, $I = 4 I_{0}$
For $ϕ = (\frac{π}{2})$ then,
$I^{'} = 4 I_{0} \cos^{2} (\frac{π}{4})$
$I^{'} = 4 I_{0} \times \frac{1}{2}$
$I^{'} = \frac{I}{2} (∵ I_{o} = \frac{I}{4})$

WB JEE 2022

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: