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PHXII10:WAVE OPTICS

368137 In Young's double slit experiment, the wavelength of red light is $7800$ $A^{\circ}$ and that of blue light is $5200$ $A^{\circ}$ . The value of $n$ for which $n^{th}$ bright band due to red light coincides with $(n + 1)^{th}$ bright band due to blue light, is

1 1

2 2

3 3

4 4

Explanation:

We have,
$y_{n} ($ Red $) = y_{n + 1} ($ Blue $)$
$\Rightarrow 7800 n = 5200 (n + 1) [∵ y_{n} = \frac{n λ D}{d}]$
$\Rightarrow n = 2$

PHXII10:WAVE OPTICS

368138 In Young’s double slit experiment, slits are separated by $2 m m$ and the screen is placed at a distance of $1.2 m$ from the slits. Light consisting of two wavelengths $6500 A^{\circ}$ and $5200 A^{\circ}$ are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

1

0.312 m m

2

0.123 m m

3

0.213 m m

4

0.412 m m

Explanation:

Distance between slits, $d = 2 m m = 2 \times 10^{- 3} m$
Distance of the screen from the slits, $D = 1.2 m$
Wavelengths,
$λ_{1} = 6500 \overset{\circ}{A} = 6500 \times 10^{- 10} = 6.5 \times 10^{- 7} m$
and $λ_{2} = 5200 \overset{\circ}{A} = 5200 \times 10^{- 10} = 5.2 \times 10^{- 7} m$
Distance of $n^{t h}$ bright fringe from the centre
$x_{n} = \frac{n λ D}{d}$
Thus, the separation between the ${4^{t}}^{h}$ bright fringes due to $λ_{1} a n d λ_{2},$
$Δ x = \frac{4 λ_{1} D}{d} - \frac{4 λ_{2} D}{d} = \frac{4 D (λ_{1} - λ_{2})}{d}$
$= \frac{4 (1.2 m) (6.5 \times 1 0^{- 7} m - 5.2 \times 1 0^{- 7} m)}{(2 \times 1 0^{- 3} m)}$
$= 3.12 \times 10^{- 4} m = 0.312 m m$

KCET - 2018

PHXII10:WAVE OPTICS

368139 The Young's double slit experiment is performed with blue and green light of wavelengths $4360 \overset{^{\circ}}{A}$ and $5460 \overset{^{\circ}}{A}$ respectively. If $x$ is the distance of 4th maxima from the central one, then

1

x_{b l u e} = x_{g r e e n}

2

x_{b l u e} > x_{g r e e n}

3

x_{b l u e} < x_{g r e e n}

4

x_{b l u e} / x_{g r e e n}

Explanation:

In Young's double slit experiment position of bright fringes on the screen are given by
$x_{n} = \frac{n D λ}{d}$
Hence, distance of $n t h$ maxima will be
$x_{n} = n λ \frac{D}{d} \propto λ$
$λ_{b l u e} = 4360 \overset{^{\circ}}{A}, λ_{g r e e n} = 5480 \overset{^{\circ}}{A}$
Here, $x =$ distance of $4^{th}$ maxima from the central one
$4 = x$ (given)
As, $λ_{b l u e} < λ_{g r e e n} ∴ x_{b l u e} < x_{g r e e n}$
$(f o r 4^{t h} m a x i m a)$

AIIMS - 2017

PHXII10:WAVE OPTICS

368140 In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

1 There shall be an interference pattern for red distinct from that for blue

2 There shall be alternative interference patterns of red and blue

3 There shall be an interference pattern for red mixing with one for blue

4 There shall be no interference fringes

Explanation:

As light from two slits of $Y D S E$ is of different colours/wavelength/frequencies, therefore, there shall be no interference fringes.

NCERT Exemplar

PHXII10:WAVE OPTICS

368137 In Young's double slit experiment, the wavelength of red light is $7800$ $A^{\circ}$ and that of blue light is $5200$ $A^{\circ}$ . The value of $n$ for which $n^{th}$ bright band due to red light coincides with $(n + 1)^{th}$ bright band due to blue light, is

1 1

2 2

3 3

4 4

Explanation:

We have,
$y_{n} ($ Red $) = y_{n + 1} ($ Blue $)$
$\Rightarrow 7800 n = 5200 (n + 1) [∵ y_{n} = \frac{n λ D}{d}]$
$\Rightarrow n = 2$

PHXII10:WAVE OPTICS

368138 In Young’s double slit experiment, slits are separated by $2 m m$ and the screen is placed at a distance of $1.2 m$ from the slits. Light consisting of two wavelengths $6500 A^{\circ}$ and $5200 A^{\circ}$ are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

1

0.312 m m

2

0.123 m m

3

0.213 m m

4

0.412 m m

Explanation:

Distance between slits, $d = 2 m m = 2 \times 10^{- 3} m$
Distance of the screen from the slits, $D = 1.2 m$
Wavelengths,
$λ_{1} = 6500 \overset{\circ}{A} = 6500 \times 10^{- 10} = 6.5 \times 10^{- 7} m$
and $λ_{2} = 5200 \overset{\circ}{A} = 5200 \times 10^{- 10} = 5.2 \times 10^{- 7} m$
Distance of $n^{t h}$ bright fringe from the centre
$x_{n} = \frac{n λ D}{d}$
Thus, the separation between the ${4^{t}}^{h}$ bright fringes due to $λ_{1} a n d λ_{2},$
$Δ x = \frac{4 λ_{1} D}{d} - \frac{4 λ_{2} D}{d} = \frac{4 D (λ_{1} - λ_{2})}{d}$
$= \frac{4 (1.2 m) (6.5 \times 1 0^{- 7} m - 5.2 \times 1 0^{- 7} m)}{(2 \times 1 0^{- 3} m)}$
$= 3.12 \times 10^{- 4} m = 0.312 m m$

KCET - 2018

PHXII10:WAVE OPTICS

368139 The Young's double slit experiment is performed with blue and green light of wavelengths $4360 \overset{^{\circ}}{A}$ and $5460 \overset{^{\circ}}{A}$ respectively. If $x$ is the distance of 4th maxima from the central one, then

1

x_{b l u e} = x_{g r e e n}

2

x_{b l u e} > x_{g r e e n}

3

x_{b l u e} < x_{g r e e n}

4

x_{b l u e} / x_{g r e e n}

Explanation:

In Young's double slit experiment position of bright fringes on the screen are given by
$x_{n} = \frac{n D λ}{d}$
Hence, distance of $n t h$ maxima will be
$x_{n} = n λ \frac{D}{d} \propto λ$
$λ_{b l u e} = 4360 \overset{^{\circ}}{A}, λ_{g r e e n} = 5480 \overset{^{\circ}}{A}$
Here, $x =$ distance of $4^{th}$ maxima from the central one
$4 = x$ (given)
As, $λ_{b l u e} < λ_{g r e e n} ∴ x_{b l u e} < x_{g r e e n}$
$(f o r 4^{t h} m a x i m a)$

AIIMS - 2017

PHXII10:WAVE OPTICS

368140 In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

1 There shall be an interference pattern for red distinct from that for blue

2 There shall be alternative interference patterns of red and blue

3 There shall be an interference pattern for red mixing with one for blue

4 There shall be no interference fringes

Explanation:

As light from two slits of $Y D S E$ is of different colours/wavelength/frequencies, therefore, there shall be no interference fringes.

NCERT Exemplar

PHXII10:WAVE OPTICS

368137 In Young's double slit experiment, the wavelength of red light is $7800$ $A^{\circ}$ and that of blue light is $5200$ $A^{\circ}$ . The value of $n$ for which $n^{th}$ bright band due to red light coincides with $(n + 1)^{th}$ bright band due to blue light, is

1 1

2 2

3 3

4 4

Explanation:

We have,
$y_{n} ($ Red $) = y_{n + 1} ($ Blue $)$
$\Rightarrow 7800 n = 5200 (n + 1) [∵ y_{n} = \frac{n λ D}{d}]$
$\Rightarrow n = 2$

PHXII10:WAVE OPTICS

368138 In Young’s double slit experiment, slits are separated by $2 m m$ and the screen is placed at a distance of $1.2 m$ from the slits. Light consisting of two wavelengths $6500 A^{\circ}$ and $5200 A^{\circ}$ are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

1

0.312 m m

2

0.123 m m

3

0.213 m m

4

0.412 m m

Explanation:

Distance between slits, $d = 2 m m = 2 \times 10^{- 3} m$
Distance of the screen from the slits, $D = 1.2 m$
Wavelengths,
$λ_{1} = 6500 \overset{\circ}{A} = 6500 \times 10^{- 10} = 6.5 \times 10^{- 7} m$
and $λ_{2} = 5200 \overset{\circ}{A} = 5200 \times 10^{- 10} = 5.2 \times 10^{- 7} m$
Distance of $n^{t h}$ bright fringe from the centre
$x_{n} = \frac{n λ D}{d}$
Thus, the separation between the ${4^{t}}^{h}$ bright fringes due to $λ_{1} a n d λ_{2},$
$Δ x = \frac{4 λ_{1} D}{d} - \frac{4 λ_{2} D}{d} = \frac{4 D (λ_{1} - λ_{2})}{d}$
$= \frac{4 (1.2 m) (6.5 \times 1 0^{- 7} m - 5.2 \times 1 0^{- 7} m)}{(2 \times 1 0^{- 3} m)}$
$= 3.12 \times 10^{- 4} m = 0.312 m m$

KCET - 2018

PHXII10:WAVE OPTICS

368139 The Young's double slit experiment is performed with blue and green light of wavelengths $4360 \overset{^{\circ}}{A}$ and $5460 \overset{^{\circ}}{A}$ respectively. If $x$ is the distance of 4th maxima from the central one, then

1

x_{b l u e} = x_{g r e e n}

2

x_{b l u e} > x_{g r e e n}

3

x_{b l u e} < x_{g r e e n}

4

x_{b l u e} / x_{g r e e n}

Explanation:

In Young's double slit experiment position of bright fringes on the screen are given by
$x_{n} = \frac{n D λ}{d}$
Hence, distance of $n t h$ maxima will be
$x_{n} = n λ \frac{D}{d} \propto λ$
$λ_{b l u e} = 4360 \overset{^{\circ}}{A}, λ_{g r e e n} = 5480 \overset{^{\circ}}{A}$
Here, $x =$ distance of $4^{th}$ maxima from the central one
$4 = x$ (given)
As, $λ_{b l u e} < λ_{g r e e n} ∴ x_{b l u e} < x_{g r e e n}$
$(f o r 4^{t h} m a x i m a)$

AIIMS - 2017

PHXII10:WAVE OPTICS

368140 In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

1 There shall be an interference pattern for red distinct from that for blue

2 There shall be alternative interference patterns of red and blue

3 There shall be an interference pattern for red mixing with one for blue

4 There shall be no interference fringes

Explanation:

As light from two slits of $Y D S E$ is of different colours/wavelength/frequencies, therefore, there shall be no interference fringes.

NCERT Exemplar

PHXII10:WAVE OPTICS

368137 In Young's double slit experiment, the wavelength of red light is $7800$ $A^{\circ}$ and that of blue light is $5200$ $A^{\circ}$ . The value of $n$ for which $n^{th}$ bright band due to red light coincides with $(n + 1)^{th}$ bright band due to blue light, is

1 1

2 2

3 3

4 4

Explanation:

We have,
$y_{n} ($ Red $) = y_{n + 1} ($ Blue $)$
$\Rightarrow 7800 n = 5200 (n + 1) [∵ y_{n} = \frac{n λ D}{d}]$
$\Rightarrow n = 2$

PHXII10:WAVE OPTICS

368138 In Young’s double slit experiment, slits are separated by $2 m m$ and the screen is placed at a distance of $1.2 m$ from the slits. Light consisting of two wavelengths $6500 A^{\circ}$ and $5200 A^{\circ}$ are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is

1

0.312 m m

2

0.123 m m

3

0.213 m m

4

0.412 m m

Explanation:

Distance between slits, $d = 2 m m = 2 \times 10^{- 3} m$
Distance of the screen from the slits, $D = 1.2 m$
Wavelengths,
$λ_{1} = 6500 \overset{\circ}{A} = 6500 \times 10^{- 10} = 6.5 \times 10^{- 7} m$
and $λ_{2} = 5200 \overset{\circ}{A} = 5200 \times 10^{- 10} = 5.2 \times 10^{- 7} m$
Distance of $n^{t h}$ bright fringe from the centre
$x_{n} = \frac{n λ D}{d}$
Thus, the separation between the ${4^{t}}^{h}$ bright fringes due to $λ_{1} a n d λ_{2},$
$Δ x = \frac{4 λ_{1} D}{d} - \frac{4 λ_{2} D}{d} = \frac{4 D (λ_{1} - λ_{2})}{d}$
$= \frac{4 (1.2 m) (6.5 \times 1 0^{- 7} m - 5.2 \times 1 0^{- 7} m)}{(2 \times 1 0^{- 3} m)}$
$= 3.12 \times 10^{- 4} m = 0.312 m m$

KCET - 2018

PHXII10:WAVE OPTICS

368139 The Young's double slit experiment is performed with blue and green light of wavelengths $4360 \overset{^{\circ}}{A}$ and $5460 \overset{^{\circ}}{A}$ respectively. If $x$ is the distance of 4th maxima from the central one, then

1

x_{b l u e} = x_{g r e e n}

2

x_{b l u e} > x_{g r e e n}

3

x_{b l u e} < x_{g r e e n}

4

x_{b l u e} / x_{g r e e n}

Explanation:

In Young's double slit experiment position of bright fringes on the screen are given by
$x_{n} = \frac{n D λ}{d}$
Hence, distance of $n t h$ maxima will be
$x_{n} = n λ \frac{D}{d} \propto λ$
$λ_{b l u e} = 4360 \overset{^{\circ}}{A}, λ_{g r e e n} = 5480 \overset{^{\circ}}{A}$
Here, $x =$ distance of $4^{th}$ maxima from the central one
$4 = x$ (given)
As, $λ_{b l u e} < λ_{g r e e n} ∴ x_{b l u e} < x_{g r e e n}$
$(f o r 4^{t h} m a x i m a)$

AIIMS - 2017

PHXII10:WAVE OPTICS

368140 In a Young’s double slit experiment, the source is white light. One of the holes is covered by a red filter and another by a blue filter. In this case

1 There shall be an interference pattern for red distinct from that for blue

2 There shall be alternative interference patterns of red and blue

3 There shall be an interference pattern for red mixing with one for blue

4 There shall be no interference fringes

Explanation:

As light from two slits of $Y D S E$ is of different colours/wavelength/frequencies, therefore, there shall be no interference fringes.

NCERT Exemplar