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PHXII10:WAVE OPTICS

368033 In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000 A^{\circ}$ . If the angular width of a fringe formed on a distant screen is $1^{\circ}$ , then value of $d$ is

1

1 m m

2

0.05 m m

3

0.03 m m

4

0.01 m m

Explanation:

Angular fringe width, $θ = \frac{λ}{d}$
or, $d = \frac{λ}{θ}$
Here, $λ = 6000 \overset{^{\circ}}{A} = 6000 \times 10^{- 10} m$
$\begin{aligned} θ = 1^{\circ} = \frac{π}{180} rad \\ ∴ d = \frac{6000 \times 10^{- 10}}{π / 180} \end{aligned}$
$= 3.44 \times 10^{- 5} m = 0.03 m m$

PHXII10:WAVE OPTICS

368034 In Young’s double slit experiment using monochromatic light of wavelength $λ$ , the intensity of light at a point on the screen with path difference $λ$ is $M$ unit. The intensity of light at a point where path difference is $λ / 3$ is

1

\frac{M}{8}

2

\frac{M}{16}

3

\frac{M}{4}

4

\frac{M}{2}

Explanation:

$I_{R} = 4 I \cos^{2} (\frac{ϕ}{2})$
$ϕ = λ \times \frac{2 π}{λ} = 2 π$
$I_{R} = 4 I = M$
$I f ϕ = \frac{λ}{3} \times \frac{2 π}{λ} = \frac{2 π}{3} r a d$
${I_{R}}^{'} = 4 I \cos^{2} (\frac{2 π}{3}) = I = \frac{M}{4}$

PHXII10:WAVE OPTICS

368035 The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensities in the interference pattern is

1

4 : 1

2

1 : 1

3

9 : 1

4

16 : 1

Explanation:

Intensity $\propto$ width of slit
If $I_{1} = I$
$∴ I_{2} = 4 I$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (\sqrt{I} + 2 \sqrt{I})^{2} = 9 I$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (\sqrt{I} - 2 \sqrt{I})^{2} = I$
so, $\frac{I_{max}}{I_{min}} = 9$
Therefore, correct option is (3).

JEE - 2024

PHXII10:WAVE OPTICS

368036 In a Young’s double slit experiment, the two slits act as coherent sources of wave of equal amplitude $A$ and wavelength $λ$ . In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I_{1}$ and in the second case is $I_{2}$ then the ratio $\frac{I_{1}}{I_{2}}$ is

1 2

2 1

3 0.5

4 4

Explanation:

For coherent sources:
$I_{1} = 4 I$ ,
For incoherent sources
$I_{2} = 2 I ∴ \frac{I_{1}}{I_{2}} = \frac{2}{1}$

PHXII10:WAVE OPTICS

368033 In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000 A^{\circ}$ . If the angular width of a fringe formed on a distant screen is $1^{\circ}$ , then value of $d$ is

1

1 m m

2

0.05 m m

3

0.03 m m

4

0.01 m m

Explanation:

Angular fringe width, $θ = \frac{λ}{d}$
or, $d = \frac{λ}{θ}$
Here, $λ = 6000 \overset{^{\circ}}{A} = 6000 \times 10^{- 10} m$
$\begin{aligned} θ = 1^{\circ} = \frac{π}{180} rad \\ ∴ d = \frac{6000 \times 10^{- 10}}{π / 180} \end{aligned}$
$= 3.44 \times 10^{- 5} m = 0.03 m m$

PHXII10:WAVE OPTICS

368034 In Young’s double slit experiment using monochromatic light of wavelength $λ$ , the intensity of light at a point on the screen with path difference $λ$ is $M$ unit. The intensity of light at a point where path difference is $λ / 3$ is

1

\frac{M}{8}

2

\frac{M}{16}

3

\frac{M}{4}

4

\frac{M}{2}

Explanation:

$I_{R} = 4 I \cos^{2} (\frac{ϕ}{2})$
$ϕ = λ \times \frac{2 π}{λ} = 2 π$
$I_{R} = 4 I = M$
$I f ϕ = \frac{λ}{3} \times \frac{2 π}{λ} = \frac{2 π}{3} r a d$
${I_{R}}^{'} = 4 I \cos^{2} (\frac{2 π}{3}) = I = \frac{M}{4}$

PHXII10:WAVE OPTICS

368035 The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensities in the interference pattern is

1

4 : 1

2

1 : 1

3

9 : 1

4

16 : 1

Explanation:

Intensity $\propto$ width of slit
If $I_{1} = I$
$∴ I_{2} = 4 I$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (\sqrt{I} + 2 \sqrt{I})^{2} = 9 I$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (\sqrt{I} - 2 \sqrt{I})^{2} = I$
so, $\frac{I_{max}}{I_{min}} = 9$
Therefore, correct option is (3).

JEE - 2024

PHXII10:WAVE OPTICS

368036 In a Young’s double slit experiment, the two slits act as coherent sources of wave of equal amplitude $A$ and wavelength $λ$ . In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I_{1}$ and in the second case is $I_{2}$ then the ratio $\frac{I_{1}}{I_{2}}$ is

1 2

2 1

3 0.5

4 4

Explanation:

For coherent sources:
$I_{1} = 4 I$ ,
For incoherent sources
$I_{2} = 2 I ∴ \frac{I_{1}}{I_{2}} = \frac{2}{1}$

PHXII10:WAVE OPTICS

368033 In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000 A^{\circ}$ . If the angular width of a fringe formed on a distant screen is $1^{\circ}$ , then value of $d$ is

1

1 m m

2

0.05 m m

3

0.03 m m

4

0.01 m m

Explanation:

Angular fringe width, $θ = \frac{λ}{d}$
or, $d = \frac{λ}{θ}$
Here, $λ = 6000 \overset{^{\circ}}{A} = 6000 \times 10^{- 10} m$
$\begin{aligned} θ = 1^{\circ} = \frac{π}{180} rad \\ ∴ d = \frac{6000 \times 10^{- 10}}{π / 180} \end{aligned}$
$= 3.44 \times 10^{- 5} m = 0.03 m m$

PHXII10:WAVE OPTICS

368034 In Young’s double slit experiment using monochromatic light of wavelength $λ$ , the intensity of light at a point on the screen with path difference $λ$ is $M$ unit. The intensity of light at a point where path difference is $λ / 3$ is

1

\frac{M}{8}

2

\frac{M}{16}

3

\frac{M}{4}

4

\frac{M}{2}

Explanation:

$I_{R} = 4 I \cos^{2} (\frac{ϕ}{2})$
$ϕ = λ \times \frac{2 π}{λ} = 2 π$
$I_{R} = 4 I = M$
$I f ϕ = \frac{λ}{3} \times \frac{2 π}{λ} = \frac{2 π}{3} r a d$
${I_{R}}^{'} = 4 I \cos^{2} (\frac{2 π}{3}) = I = \frac{M}{4}$

PHXII10:WAVE OPTICS

368035 The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensities in the interference pattern is

1

4 : 1

2

1 : 1

3

9 : 1

4

16 : 1

Explanation:

Intensity $\propto$ width of slit
If $I_{1} = I$
$∴ I_{2} = 4 I$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (\sqrt{I} + 2 \sqrt{I})^{2} = 9 I$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (\sqrt{I} - 2 \sqrt{I})^{2} = I$
so, $\frac{I_{max}}{I_{min}} = 9$
Therefore, correct option is (3).

JEE - 2024

PHXII10:WAVE OPTICS

368036 In a Young’s double slit experiment, the two slits act as coherent sources of wave of equal amplitude $A$ and wavelength $λ$ . In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I_{1}$ and in the second case is $I_{2}$ then the ratio $\frac{I_{1}}{I_{2}}$ is

1 2

2 1

3 0.5

4 4

Explanation:

For coherent sources:
$I_{1} = 4 I$ ,
For incoherent sources
$I_{2} = 2 I ∴ \frac{I_{1}}{I_{2}} = \frac{2}{1}$

PHXII10:WAVE OPTICS

368033 In Young's double slit experiment, the spacing between the slits is $d$ and wavelength of light used is $6000 A^{\circ}$ . If the angular width of a fringe formed on a distant screen is $1^{\circ}$ , then value of $d$ is

1

1 m m

2

0.05 m m

3

0.03 m m

4

0.01 m m

Explanation:

Angular fringe width, $θ = \frac{λ}{d}$
or, $d = \frac{λ}{θ}$
Here, $λ = 6000 \overset{^{\circ}}{A} = 6000 \times 10^{- 10} m$
$\begin{aligned} θ = 1^{\circ} = \frac{π}{180} rad \\ ∴ d = \frac{6000 \times 10^{- 10}}{π / 180} \end{aligned}$
$= 3.44 \times 10^{- 5} m = 0.03 m m$

PHXII10:WAVE OPTICS

368034 In Young’s double slit experiment using monochromatic light of wavelength $λ$ , the intensity of light at a point on the screen with path difference $λ$ is $M$ unit. The intensity of light at a point where path difference is $λ / 3$ is

1

\frac{M}{8}

2

\frac{M}{16}

3

\frac{M}{4}

4

\frac{M}{2}

Explanation:

$I_{R} = 4 I \cos^{2} (\frac{ϕ}{2})$
$ϕ = λ \times \frac{2 π}{λ} = 2 π$
$I_{R} = 4 I = M$
$I f ϕ = \frac{λ}{3} \times \frac{2 π}{λ} = \frac{2 π}{3} r a d$
${I_{R}}^{'} = 4 I \cos^{2} (\frac{2 π}{3}) = I = \frac{M}{4}$

PHXII10:WAVE OPTICS

368035 The width of one of the two slits in a Young's double slit experiment is 4 times that of the other slit. The ratio of the maximum to the minimum intensities in the interference pattern is

1

4 : 1

2

1 : 1

3

9 : 1

4

16 : 1

Explanation:

Intensity $\propto$ width of slit
If $I_{1} = I$
$∴ I_{2} = 4 I$
$I_{max} = {(\sqrt{I_{1}} + \sqrt{I_{2}})}^{2} = (\sqrt{I} + 2 \sqrt{I})^{2} = 9 I$
$I_{min} = {(\sqrt{I_{1}} - \sqrt{I_{2}})}^{2} = (\sqrt{I} - 2 \sqrt{I})^{2} = I$
so, $\frac{I_{max}}{I_{min}} = 9$
Therefore, correct option is (3).

JEE - 2024

PHXII10:WAVE OPTICS

368036 In a Young’s double slit experiment, the two slits act as coherent sources of wave of equal amplitude $A$ and wavelength $λ$ . In another experiment with the same arrangement the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I_{1}$ and in the second case is $I_{2}$ then the ratio $\frac{I_{1}}{I_{2}}$ is

1 2

2 1

3 0.5

4 4

Explanation:

For coherent sources:
$I_{1} = 4 I$ ,
For incoherent sources
$I_{2} = 2 I ∴ \frac{I_{1}}{I_{2}} = \frac{2}{1}$