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WAVE OPTICS

283354 In Young's double slit experiment, the fringe width is found to be $0.4 mm$ . If the whole apparatus is dipped in water of refractive index $4 / 3$ , without disturbing the arrangement, the new fringe width will be

1

0.30 mm

2

0.40 mm

3

0.53 mm

4

0.2 mm

Explanation:

: Given,
$β = 0.4 mm$
We know,
$Fringe width (β) = \frac{D λ}{d}$
When whole apparatus is immersed in liquid then ' $D$ ' and 'd' will not change only ' $λ$ ' will be changeNow,
$λ^{'} = \frac{3}{4} λ$
So,
$β \propto λ$
$\frac{β}{β^{'}} = \frac{λ}{λ^{'}}$
$\frac{0.4}{β^{'}} = \frac{λ}{\frac{3}{4} λ}$
$β^{'} = \frac{3 \times 0.4}{4}$
$β^{'} = 0.30 mm$

Assam CEE-2017

WAVE OPTICS

283355 In Young's double slit experiment the slits are horizontal. The intensity at a point ' $P$ ' on the screen shown in the figure is $\frac{I_{0}}{4}$ where $I_{0}$ is maximum intensity. If the distance between the two slits $S_{1}$ and $S$ , is $2 λ$ . then the value of ' $θ$ ' is

1

\cos^{- 1} (\frac{1}{6})

2

\sin^{- 1} (\frac{1}{12})

3

\tan^{- 1} (\frac{1}{6})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

: $I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
original image
Phase difference $(ϕ) = \frac{2 π}{λ} \times$ path difference $(Δ x)$
$\frac{2 π}{3} = \frac{2 π}{λ} \times d \cos θ$
$\frac{1}{3} = \frac{1}{λ} \times 2 λ \cos θ$
$\cos θ = \frac{1}{6}$
$θ = \cos^{- 1} (\frac{1}{6})$

AP EAMCET-25.04.2017

WAVE OPTICS

283356 In Young's double slit experiment, intensity at a point on the screen is $\frac{1}{4}$ th of the maximum intensity. Then the angular position of this point is $(λ$ - wavelength of light. $d$ - separation between the two slits)

1

\sin^{- 1} (\frac{λ}{d})

2

\sin^{- 1} (\frac{λ}{3 d})

3

\sin^{- 1} (\frac{3 λ}{2 d})

4

\tan^{- 1} (\frac{λ}{d})

Explanation:

: $I = \frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
$ϕ = \frac{2 π}{λ} \times Δ x$
$\frac{2 π}{3} = \frac{2 π}{λ} \times Δ x$
$Δ x = \frac{λ}{3}$
We know that
$Δ x = d \sin θ$
$\sin θ = \frac{Δ x}{d} = \frac{λ}{3 d}$
$θ = \sin^{- 1} (\frac{λ}{3 d})$

MHT-CET 2020

WAVE OPTICS

283357 In Young's double slit experiment, the slits separated by $0.6 mm$ are illuminated with light of $6600 \AA$ . Interference pattern is obtained on a screen placed at $4 m$ from slits. The minimum distance from the central maximum at which the average intensity is $50 %$ of the maximum value is

1

0.21 mm

2

2.1 mm

3

0.11 mm

4

1.1 mm

Explanation:

: Given that,
$λ = 6600 \AA = 6600 \times 10^{- 8} m$
$d = 0.6 mm = 0.6 \times 10^{- 3} m$
$D = 4 m$
And, $I = 50 %$ of $I_{0}$ .
$I = \frac{I_{0}}{2}$
We know that,
$I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{1}{2} [\frac{I_{0}}{2}] = \frac{I_{0}}{2} \cos^{2} \frac{ϕ}{2}$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{2}$
$\frac{ϕ}{2} = \frac{π}{4}$
$ϕ = \frac{π}{2}$
Path difference $(Δ x) = \frac{λ}{2 π} ϕ$
$\frac{yd}{D} = \frac{λ}{2 π} \times \frac{π}{2}$
$\frac{yd}{D} = \frac{λ}{4}$
$y = \frac{λ D}{4 d}$
$= \frac{66 \times 10^{- 8} \times 4}{4 \times 0.6 \times 10^{- 3}}$
$= 11 \times 10^{- 4} m$
$= 1.1 mm$

AP EAMCET-26.04.2017

WAVE OPTICS

283354 In Young's double slit experiment, the fringe width is found to be $0.4 mm$ . If the whole apparatus is dipped in water of refractive index $4 / 3$ , without disturbing the arrangement, the new fringe width will be

1

0.30 mm

2

0.40 mm

3

0.53 mm

4

0.2 mm

Explanation:

: Given,
$β = 0.4 mm$
We know,
$Fringe width (β) = \frac{D λ}{d}$
When whole apparatus is immersed in liquid then ' $D$ ' and 'd' will not change only ' $λ$ ' will be changeNow,
$λ^{'} = \frac{3}{4} λ$
So,
$β \propto λ$
$\frac{β}{β^{'}} = \frac{λ}{λ^{'}}$
$\frac{0.4}{β^{'}} = \frac{λ}{\frac{3}{4} λ}$
$β^{'} = \frac{3 \times 0.4}{4}$
$β^{'} = 0.30 mm$

Assam CEE-2017

WAVE OPTICS

283355 In Young's double slit experiment the slits are horizontal. The intensity at a point ' $P$ ' on the screen shown in the figure is $\frac{I_{0}}{4}$ where $I_{0}$ is maximum intensity. If the distance between the two slits $S_{1}$ and $S$ , is $2 λ$ . then the value of ' $θ$ ' is

1

\cos^{- 1} (\frac{1}{6})

2

\sin^{- 1} (\frac{1}{12})

3

\tan^{- 1} (\frac{1}{6})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

: $I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
original image
Phase difference $(ϕ) = \frac{2 π}{λ} \times$ path difference $(Δ x)$
$\frac{2 π}{3} = \frac{2 π}{λ} \times d \cos θ$
$\frac{1}{3} = \frac{1}{λ} \times 2 λ \cos θ$
$\cos θ = \frac{1}{6}$
$θ = \cos^{- 1} (\frac{1}{6})$

AP EAMCET-25.04.2017

WAVE OPTICS

283356 In Young's double slit experiment, intensity at a point on the screen is $\frac{1}{4}$ th of the maximum intensity. Then the angular position of this point is $(λ$ - wavelength of light. $d$ - separation between the two slits)

1

\sin^{- 1} (\frac{λ}{d})

2

\sin^{- 1} (\frac{λ}{3 d})

3

\sin^{- 1} (\frac{3 λ}{2 d})

4

\tan^{- 1} (\frac{λ}{d})

Explanation:

: $I = \frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
$ϕ = \frac{2 π}{λ} \times Δ x$
$\frac{2 π}{3} = \frac{2 π}{λ} \times Δ x$
$Δ x = \frac{λ}{3}$
We know that
$Δ x = d \sin θ$
$\sin θ = \frac{Δ x}{d} = \frac{λ}{3 d}$
$θ = \sin^{- 1} (\frac{λ}{3 d})$

MHT-CET 2020

WAVE OPTICS

283357 In Young's double slit experiment, the slits separated by $0.6 mm$ are illuminated with light of $6600 \AA$ . Interference pattern is obtained on a screen placed at $4 m$ from slits. The minimum distance from the central maximum at which the average intensity is $50 %$ of the maximum value is

1

0.21 mm

2

2.1 mm

3

0.11 mm

4

1.1 mm

Explanation:

: Given that,
$λ = 6600 \AA = 6600 \times 10^{- 8} m$
$d = 0.6 mm = 0.6 \times 10^{- 3} m$
$D = 4 m$
And, $I = 50 %$ of $I_{0}$ .
$I = \frac{I_{0}}{2}$
We know that,
$I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{1}{2} [\frac{I_{0}}{2}] = \frac{I_{0}}{2} \cos^{2} \frac{ϕ}{2}$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{2}$
$\frac{ϕ}{2} = \frac{π}{4}$
$ϕ = \frac{π}{2}$
Path difference $(Δ x) = \frac{λ}{2 π} ϕ$
$\frac{yd}{D} = \frac{λ}{2 π} \times \frac{π}{2}$
$\frac{yd}{D} = \frac{λ}{4}$
$y = \frac{λ D}{4 d}$
$= \frac{66 \times 10^{- 8} \times 4}{4 \times 0.6 \times 10^{- 3}}$
$= 11 \times 10^{- 4} m$
$= 1.1 mm$

AP EAMCET-26.04.2017

WAVE OPTICS

283354 In Young's double slit experiment, the fringe width is found to be $0.4 mm$ . If the whole apparatus is dipped in water of refractive index $4 / 3$ , without disturbing the arrangement, the new fringe width will be

1

0.30 mm

2

0.40 mm

3

0.53 mm

4

0.2 mm

Explanation:

: Given,
$β = 0.4 mm$
We know,
$Fringe width (β) = \frac{D λ}{d}$
When whole apparatus is immersed in liquid then ' $D$ ' and 'd' will not change only ' $λ$ ' will be changeNow,
$λ^{'} = \frac{3}{4} λ$
So,
$β \propto λ$
$\frac{β}{β^{'}} = \frac{λ}{λ^{'}}$
$\frac{0.4}{β^{'}} = \frac{λ}{\frac{3}{4} λ}$
$β^{'} = \frac{3 \times 0.4}{4}$
$β^{'} = 0.30 mm$

Assam CEE-2017

WAVE OPTICS

283355 In Young's double slit experiment the slits are horizontal. The intensity at a point ' $P$ ' on the screen shown in the figure is $\frac{I_{0}}{4}$ where $I_{0}$ is maximum intensity. If the distance between the two slits $S_{1}$ and $S$ , is $2 λ$ . then the value of ' $θ$ ' is

1

\cos^{- 1} (\frac{1}{6})

2

\sin^{- 1} (\frac{1}{12})

3

\tan^{- 1} (\frac{1}{6})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

: $I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
original image
Phase difference $(ϕ) = \frac{2 π}{λ} \times$ path difference $(Δ x)$
$\frac{2 π}{3} = \frac{2 π}{λ} \times d \cos θ$
$\frac{1}{3} = \frac{1}{λ} \times 2 λ \cos θ$
$\cos θ = \frac{1}{6}$
$θ = \cos^{- 1} (\frac{1}{6})$

AP EAMCET-25.04.2017

WAVE OPTICS

283356 In Young's double slit experiment, intensity at a point on the screen is $\frac{1}{4}$ th of the maximum intensity. Then the angular position of this point is $(λ$ - wavelength of light. $d$ - separation between the two slits)

1

\sin^{- 1} (\frac{λ}{d})

2

\sin^{- 1} (\frac{λ}{3 d})

3

\sin^{- 1} (\frac{3 λ}{2 d})

4

\tan^{- 1} (\frac{λ}{d})

Explanation:

: $I = \frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
$ϕ = \frac{2 π}{λ} \times Δ x$
$\frac{2 π}{3} = \frac{2 π}{λ} \times Δ x$
$Δ x = \frac{λ}{3}$
We know that
$Δ x = d \sin θ$
$\sin θ = \frac{Δ x}{d} = \frac{λ}{3 d}$
$θ = \sin^{- 1} (\frac{λ}{3 d})$

MHT-CET 2020

WAVE OPTICS

283357 In Young's double slit experiment, the slits separated by $0.6 mm$ are illuminated with light of $6600 \AA$ . Interference pattern is obtained on a screen placed at $4 m$ from slits. The minimum distance from the central maximum at which the average intensity is $50 %$ of the maximum value is

1

0.21 mm

2

2.1 mm

3

0.11 mm

4

1.1 mm

Explanation:

: Given that,
$λ = 6600 \AA = 6600 \times 10^{- 8} m$
$d = 0.6 mm = 0.6 \times 10^{- 3} m$
$D = 4 m$
And, $I = 50 %$ of $I_{0}$ .
$I = \frac{I_{0}}{2}$
We know that,
$I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{1}{2} [\frac{I_{0}}{2}] = \frac{I_{0}}{2} \cos^{2} \frac{ϕ}{2}$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{2}$
$\frac{ϕ}{2} = \frac{π}{4}$
$ϕ = \frac{π}{2}$
Path difference $(Δ x) = \frac{λ}{2 π} ϕ$
$\frac{yd}{D} = \frac{λ}{2 π} \times \frac{π}{2}$
$\frac{yd}{D} = \frac{λ}{4}$
$y = \frac{λ D}{4 d}$
$= \frac{66 \times 10^{- 8} \times 4}{4 \times 0.6 \times 10^{- 3}}$
$= 11 \times 10^{- 4} m$
$= 1.1 mm$

AP EAMCET-26.04.2017

WAVE OPTICS

283354 In Young's double slit experiment, the fringe width is found to be $0.4 mm$ . If the whole apparatus is dipped in water of refractive index $4 / 3$ , without disturbing the arrangement, the new fringe width will be

1

0.30 mm

2

0.40 mm

3

0.53 mm

4

0.2 mm

Explanation:

: Given,
$β = 0.4 mm$
We know,
$Fringe width (β) = \frac{D λ}{d}$
When whole apparatus is immersed in liquid then ' $D$ ' and 'd' will not change only ' $λ$ ' will be changeNow,
$λ^{'} = \frac{3}{4} λ$
So,
$β \propto λ$
$\frac{β}{β^{'}} = \frac{λ}{λ^{'}}$
$\frac{0.4}{β^{'}} = \frac{λ}{\frac{3}{4} λ}$
$β^{'} = \frac{3 \times 0.4}{4}$
$β^{'} = 0.30 mm$

Assam CEE-2017

WAVE OPTICS

283355 In Young's double slit experiment the slits are horizontal. The intensity at a point ' $P$ ' on the screen shown in the figure is $\frac{I_{0}}{4}$ where $I_{0}$ is maximum intensity. If the distance between the two slits $S_{1}$ and $S$ , is $2 λ$ . then the value of ' $θ$ ' is

1

\cos^{- 1} (\frac{1}{6})

2

\sin^{- 1} (\frac{1}{12})

3

\tan^{- 1} (\frac{1}{6})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

: $I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
original image
Phase difference $(ϕ) = \frac{2 π}{λ} \times$ path difference $(Δ x)$
$\frac{2 π}{3} = \frac{2 π}{λ} \times d \cos θ$
$\frac{1}{3} = \frac{1}{λ} \times 2 λ \cos θ$
$\cos θ = \frac{1}{6}$
$θ = \cos^{- 1} (\frac{1}{6})$

AP EAMCET-25.04.2017

WAVE OPTICS

283356 In Young's double slit experiment, intensity at a point on the screen is $\frac{1}{4}$ th of the maximum intensity. Then the angular position of this point is $(λ$ - wavelength of light. $d$ - separation between the two slits)

1

\sin^{- 1} (\frac{λ}{d})

2

\sin^{- 1} (\frac{λ}{3 d})

3

\sin^{- 1} (\frac{3 λ}{2 d})

4

\tan^{- 1} (\frac{λ}{d})

Explanation:

: $I = \frac{I_{0}}{4} = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{4}$
$\frac{ϕ}{2} = \frac{π}{3}$
$ϕ = \frac{2 π}{3}$
$ϕ = \frac{2 π}{λ} \times Δ x$
$\frac{2 π}{3} = \frac{2 π}{λ} \times Δ x$
$Δ x = \frac{λ}{3}$
We know that
$Δ x = d \sin θ$
$\sin θ = \frac{Δ x}{d} = \frac{λ}{3 d}$
$θ = \sin^{- 1} (\frac{λ}{3 d})$

MHT-CET 2020

WAVE OPTICS

283357 In Young's double slit experiment, the slits separated by $0.6 mm$ are illuminated with light of $6600 \AA$ . Interference pattern is obtained on a screen placed at $4 m$ from slits. The minimum distance from the central maximum at which the average intensity is $50 %$ of the maximum value is

1

0.21 mm

2

2.1 mm

3

0.11 mm

4

1.1 mm

Explanation:

: Given that,
$λ = 6600 \AA = 6600 \times 10^{- 8} m$
$d = 0.6 mm = 0.6 \times 10^{- 3} m$
$D = 4 m$
And, $I = 50 %$ of $I_{0}$ .
$I = \frac{I_{0}}{2}$
We know that,
$I = I_{0} \cos^{2} (\frac{ϕ}{2})$
$\frac{1}{2} [\frac{I_{0}}{2}] = \frac{I_{0}}{2} \cos^{2} \frac{ϕ}{2}$
$\cos^{2} (\frac{ϕ}{2}) = \frac{1}{2}$
$\frac{ϕ}{2} = \frac{π}{4}$
$ϕ = \frac{π}{2}$
Path difference $(Δ x) = \frac{λ}{2 π} ϕ$
$\frac{yd}{D} = \frac{λ}{2 π} \times \frac{π}{2}$
$\frac{yd}{D} = \frac{λ}{4}$
$y = \frac{λ D}{4 d}$
$= \frac{66 \times 10^{- 8} \times 4}{4 \times 0.6 \times 10^{- 3}}$
$= 11 \times 10^{- 4} m$
$= 1.1 mm$

AP EAMCET-26.04.2017