QBManager

PHXII10:WAVE OPTICS

368132 A beam of light consisting of two wavelengths, $650 n m$ and $520 n m$ is used to obtain interference fringes in a Young's double-slit experiment. What is the least distance from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is $3 m m$ and the distance between the plane of the slits and screen is $150 c m$ .

1

3.4 m m

2

1.3 m m

3

5.4 m m

4

7.2 m m

Explanation:

Let $^{'} y^{'}$ be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let $n_{1}^{th}$ number of bright fringe with wavelength $λ_{1}$ coincides with $n_{2}^{th}$ number of bright fringe with wavelength $λ_{2}$ .
$∴ n_{1} β_{1} = n_{2} β_{2}$
$\Rightarrow n_{1} \frac{λ_{1} D}{d} = n_{2} \frac{λ_{2} D}{d}$
$\Rightarrow n_{1} λ_{1} = n_{2} λ_{2}$ $(1)$
Also at first position, the $n^{th}$ bright fringe of one will coincide with $(n + 1)^{th}$ bright of other.
If $λ_{2} < λ_{1}$ ,
So then $n_{2} > n_{1}$
$\Rightarrow n_{2} = n_{1} + 1$ $(2)$
Using equation (2) and equation (1),
$n_{1} λ_{1} = (n_{1} + 1) λ_{2}$
$\Rightarrow (n_{1}) (650 \times 10^{- 9}) = (n_{1} + 1) (520 \times 10^{- 9})$
$\Rightarrow 65 n_{1} = 52 n_{1} + 52$
$\Rightarrow 13 n_{1} = 52$
$\Rightarrow n_{1} = 4$
Thus, $y = n_{1} β_{1}$
$= n_{1} \frac{λ_{1} D}{d}$
$= 4 [\frac{(6.5 \times 10^{- 7}) (1.5)}{3 \times 10^{- 3}}]$
$= 1.3 \times 10^{- 3} m$
$= 1.3 m m$

PHXII10:WAVE OPTICS

368133 The source that illuminates the double - slit in 'double - slit interference experiment' emits two distinct monochromatic waves of wavelength $500 n m$ and $600 n m$ , each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one move out of this central region, the two fringe systems are gradually out of step such that maximum due to one wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $n m$ is:

1 2000

2 3000

3 1000

4 1500

Explanation:

Let $λ_{1} = 500 n m, λ_{2} = 600 n m$
When maxima of first one coincides with minima of second one
$\frac{n λ_{1} D}{d} = \frac{(2 m + 1) λ_{2} D}{2 d}$
$\frac{n}{2 m + 1} = \frac{λ_{2}}{2 λ_{1}} = \frac{600}{2 \times 500} = \frac{3}{5}$
$n = \frac{3}{5} (2 m + 1)$
when $m = 2 \Rightarrow n = 3$
Path difference $P . d$ $= n λ_{1} = 1500 n m$

JEE - 2013

PHXII10:WAVE OPTICS

368134 In a Young’s double slit experiment, slits are separated by $0.5 m m,$ and the screen is placed $150 c m$ away. A beam of light consisting of two wavelengths, $650 n m a n d 520 n m,$ is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

1

9.75 m m

2

15.6 m m

3

1.56 m m

4

7.8 m m

Explanation:

For common maxima, $n_{1} λ_{1} = n_{2} λ_{2}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{520 \times 10^{- 9}}{650 \times 10^{- 9}} = \frac{4}{5}$
$F o r λ_{1}$
$y = \frac{n_{1} λ_{1} D}{d} = \frac{4 \times 650 \times 10^{- 9} \times 1.5}{0.5 \times 10^{- 3}} = 7.8 m m$

PHXII10:WAVE OPTICS

368135 Assertion :
In Young's double slit experiment using white light, the central fringe is white, the violet coloured fringe will be seen nearest to the central fringe.
Reason :
The interference patterns due to different component colours of white light overlap throughout the screen.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII10:WAVE OPTICS

368136 In a $Y D S E$ bi-chromatic light of wavelengths 400 $n m$ and 560 $n m$ are used. The distance between the slits is 0.1 $m m$ and the distance between the plane of the slits and the screen is 1 $m$ . Find the minimum distance between two successive regions of complete darkness

1

28 m m

2

32 m m

3

23 m m

4

42 m m

Explanation:

At the area of total darkness, in double slit apparatus, minima will occur for both the wavelengths which are incident simultaneously and normally.
$(\frac{2 n + 1}{2}) λ_{1} = \frac{(2 m + 1)}{2} λ_{2}$
$\Rightarrow \frac{2 n + 1}{2 m + 1} = \frac{λ_{2}}{λ_{1}}$
$\Rightarrow \frac{2 n + 1}{2 m + 1}$ $= \frac{560}{400} = \frac{7}{5} or 10 n = 14 m + 2$
By inspection, the two solutions are
(i) If $m_{1} = 2, n_{1} = 3$ (ii) If $m_{2} = 7, n_{2} = 10$ .
$∴$ Distance $Δ S = \frac{D λ_{1}}{d} [\frac{(2 n_{2} + 1) - (2 n_{1} + 1)}{2}]$
Now putting $n_{2} = 10$ and $n_{1} = 3$
$Δ S = 4 \times 7 \times 10^{- 3} m \Rightarrow Δ S = 28 m m$

PHXII10:WAVE OPTICS

368132 A beam of light consisting of two wavelengths, $650 n m$ and $520 n m$ is used to obtain interference fringes in a Young's double-slit experiment. What is the least distance from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is $3 m m$ and the distance between the plane of the slits and screen is $150 c m$ .

1

3.4 m m

2

1.3 m m

3

5.4 m m

4

7.2 m m

Explanation:

Let $^{'} y^{'}$ be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let $n_{1}^{th}$ number of bright fringe with wavelength $λ_{1}$ coincides with $n_{2}^{th}$ number of bright fringe with wavelength $λ_{2}$ .
$∴ n_{1} β_{1} = n_{2} β_{2}$
$\Rightarrow n_{1} \frac{λ_{1} D}{d} = n_{2} \frac{λ_{2} D}{d}$
$\Rightarrow n_{1} λ_{1} = n_{2} λ_{2}$ $(1)$
Also at first position, the $n^{th}$ bright fringe of one will coincide with $(n + 1)^{th}$ bright of other.
If $λ_{2} < λ_{1}$ ,
So then $n_{2} > n_{1}$
$\Rightarrow n_{2} = n_{1} + 1$ $(2)$
Using equation (2) and equation (1),
$n_{1} λ_{1} = (n_{1} + 1) λ_{2}$
$\Rightarrow (n_{1}) (650 \times 10^{- 9}) = (n_{1} + 1) (520 \times 10^{- 9})$
$\Rightarrow 65 n_{1} = 52 n_{1} + 52$
$\Rightarrow 13 n_{1} = 52$
$\Rightarrow n_{1} = 4$
Thus, $y = n_{1} β_{1}$
$= n_{1} \frac{λ_{1} D}{d}$
$= 4 [\frac{(6.5 \times 10^{- 7}) (1.5)}{3 \times 10^{- 3}}]$
$= 1.3 \times 10^{- 3} m$
$= 1.3 m m$

PHXII10:WAVE OPTICS

368133 The source that illuminates the double - slit in 'double - slit interference experiment' emits two distinct monochromatic waves of wavelength $500 n m$ and $600 n m$ , each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one move out of this central region, the two fringe systems are gradually out of step such that maximum due to one wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $n m$ is:

1 2000

2 3000

3 1000

4 1500

Explanation:

Let $λ_{1} = 500 n m, λ_{2} = 600 n m$
When maxima of first one coincides with minima of second one
$\frac{n λ_{1} D}{d} = \frac{(2 m + 1) λ_{2} D}{2 d}$
$\frac{n}{2 m + 1} = \frac{λ_{2}}{2 λ_{1}} = \frac{600}{2 \times 500} = \frac{3}{5}$
$n = \frac{3}{5} (2 m + 1)$
when $m = 2 \Rightarrow n = 3$
Path difference $P . d$ $= n λ_{1} = 1500 n m$

JEE - 2013

PHXII10:WAVE OPTICS

368134 In a Young’s double slit experiment, slits are separated by $0.5 m m,$ and the screen is placed $150 c m$ away. A beam of light consisting of two wavelengths, $650 n m a n d 520 n m,$ is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

1

9.75 m m

2

15.6 m m

3

1.56 m m

4

7.8 m m

Explanation:

For common maxima, $n_{1} λ_{1} = n_{2} λ_{2}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{520 \times 10^{- 9}}{650 \times 10^{- 9}} = \frac{4}{5}$
$F o r λ_{1}$
$y = \frac{n_{1} λ_{1} D}{d} = \frac{4 \times 650 \times 10^{- 9} \times 1.5}{0.5 \times 10^{- 3}} = 7.8 m m$

PHXII10:WAVE OPTICS

368135 Assertion :
In Young's double slit experiment using white light, the central fringe is white, the violet coloured fringe will be seen nearest to the central fringe.
Reason :
The interference patterns due to different component colours of white light overlap throughout the screen.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII10:WAVE OPTICS

368136 In a $Y D S E$ bi-chromatic light of wavelengths 400 $n m$ and 560 $n m$ are used. The distance between the slits is 0.1 $m m$ and the distance between the plane of the slits and the screen is 1 $m$ . Find the minimum distance between two successive regions of complete darkness

1

28 m m

2

32 m m

3

23 m m

4

42 m m

Explanation:

At the area of total darkness, in double slit apparatus, minima will occur for both the wavelengths which are incident simultaneously and normally.
$(\frac{2 n + 1}{2}) λ_{1} = \frac{(2 m + 1)}{2} λ_{2}$
$\Rightarrow \frac{2 n + 1}{2 m + 1} = \frac{λ_{2}}{λ_{1}}$
$\Rightarrow \frac{2 n + 1}{2 m + 1}$ $= \frac{560}{400} = \frac{7}{5} or 10 n = 14 m + 2$
By inspection, the two solutions are
(i) If $m_{1} = 2, n_{1} = 3$ (ii) If $m_{2} = 7, n_{2} = 10$ .
$∴$ Distance $Δ S = \frac{D λ_{1}}{d} [\frac{(2 n_{2} + 1) - (2 n_{1} + 1)}{2}]$
Now putting $n_{2} = 10$ and $n_{1} = 3$
$Δ S = 4 \times 7 \times 10^{- 3} m \Rightarrow Δ S = 28 m m$

PHXII10:WAVE OPTICS

368132 A beam of light consisting of two wavelengths, $650 n m$ and $520 n m$ is used to obtain interference fringes in a Young's double-slit experiment. What is the least distance from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is $3 m m$ and the distance between the plane of the slits and screen is $150 c m$ .

1

3.4 m m

2

1.3 m m

3

5.4 m m

4

7.2 m m

Explanation:

Let $^{'} y^{'}$ be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let $n_{1}^{th}$ number of bright fringe with wavelength $λ_{1}$ coincides with $n_{2}^{th}$ number of bright fringe with wavelength $λ_{2}$ .
$∴ n_{1} β_{1} = n_{2} β_{2}$
$\Rightarrow n_{1} \frac{λ_{1} D}{d} = n_{2} \frac{λ_{2} D}{d}$
$\Rightarrow n_{1} λ_{1} = n_{2} λ_{2}$ $(1)$
Also at first position, the $n^{th}$ bright fringe of one will coincide with $(n + 1)^{th}$ bright of other.
If $λ_{2} < λ_{1}$ ,
So then $n_{2} > n_{1}$
$\Rightarrow n_{2} = n_{1} + 1$ $(2)$
Using equation (2) and equation (1),
$n_{1} λ_{1} = (n_{1} + 1) λ_{2}$
$\Rightarrow (n_{1}) (650 \times 10^{- 9}) = (n_{1} + 1) (520 \times 10^{- 9})$
$\Rightarrow 65 n_{1} = 52 n_{1} + 52$
$\Rightarrow 13 n_{1} = 52$
$\Rightarrow n_{1} = 4$
Thus, $y = n_{1} β_{1}$
$= n_{1} \frac{λ_{1} D}{d}$
$= 4 [\frac{(6.5 \times 10^{- 7}) (1.5)}{3 \times 10^{- 3}}]$
$= 1.3 \times 10^{- 3} m$
$= 1.3 m m$

PHXII10:WAVE OPTICS

368133 The source that illuminates the double - slit in 'double - slit interference experiment' emits two distinct monochromatic waves of wavelength $500 n m$ and $600 n m$ , each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one move out of this central region, the two fringe systems are gradually out of step such that maximum due to one wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $n m$ is:

1 2000

2 3000

3 1000

4 1500

Explanation:

Let $λ_{1} = 500 n m, λ_{2} = 600 n m$
When maxima of first one coincides with minima of second one
$\frac{n λ_{1} D}{d} = \frac{(2 m + 1) λ_{2} D}{2 d}$
$\frac{n}{2 m + 1} = \frac{λ_{2}}{2 λ_{1}} = \frac{600}{2 \times 500} = \frac{3}{5}$
$n = \frac{3}{5} (2 m + 1)$
when $m = 2 \Rightarrow n = 3$
Path difference $P . d$ $= n λ_{1} = 1500 n m$

JEE - 2013

PHXII10:WAVE OPTICS

368134 In a Young’s double slit experiment, slits are separated by $0.5 m m,$ and the screen is placed $150 c m$ away. A beam of light consisting of two wavelengths, $650 n m a n d 520 n m,$ is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

1

9.75 m m

2

15.6 m m

3

1.56 m m

4

7.8 m m

Explanation:

For common maxima, $n_{1} λ_{1} = n_{2} λ_{2}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{520 \times 10^{- 9}}{650 \times 10^{- 9}} = \frac{4}{5}$
$F o r λ_{1}$
$y = \frac{n_{1} λ_{1} D}{d} = \frac{4 \times 650 \times 10^{- 9} \times 1.5}{0.5 \times 10^{- 3}} = 7.8 m m$

PHXII10:WAVE OPTICS

368135 Assertion :
In Young's double slit experiment using white light, the central fringe is white, the violet coloured fringe will be seen nearest to the central fringe.
Reason :
The interference patterns due to different component colours of white light overlap throughout the screen.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII10:WAVE OPTICS

368136 In a $Y D S E$ bi-chromatic light of wavelengths 400 $n m$ and 560 $n m$ are used. The distance between the slits is 0.1 $m m$ and the distance between the plane of the slits and the screen is 1 $m$ . Find the minimum distance between two successive regions of complete darkness

1

28 m m

2

32 m m

3

23 m m

4

42 m m

Explanation:

At the area of total darkness, in double slit apparatus, minima will occur for both the wavelengths which are incident simultaneously and normally.
$(\frac{2 n + 1}{2}) λ_{1} = \frac{(2 m + 1)}{2} λ_{2}$
$\Rightarrow \frac{2 n + 1}{2 m + 1} = \frac{λ_{2}}{λ_{1}}$
$\Rightarrow \frac{2 n + 1}{2 m + 1}$ $= \frac{560}{400} = \frac{7}{5} or 10 n = 14 m + 2$
By inspection, the two solutions are
(i) If $m_{1} = 2, n_{1} = 3$ (ii) If $m_{2} = 7, n_{2} = 10$ .
$∴$ Distance $Δ S = \frac{D λ_{1}}{d} [\frac{(2 n_{2} + 1) - (2 n_{1} + 1)}{2}]$
Now putting $n_{2} = 10$ and $n_{1} = 3$
$Δ S = 4 \times 7 \times 10^{- 3} m \Rightarrow Δ S = 28 m m$

PHXII10:WAVE OPTICS

368132 A beam of light consisting of two wavelengths, $650 n m$ and $520 n m$ is used to obtain interference fringes in a Young's double-slit experiment. What is the least distance from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is $3 m m$ and the distance between the plane of the slits and screen is $150 c m$ .

1

3.4 m m

2

1.3 m m

3

5.4 m m

4

7.2 m m

Explanation:

Let $^{'} y^{'}$ be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let $n_{1}^{th}$ number of bright fringe with wavelength $λ_{1}$ coincides with $n_{2}^{th}$ number of bright fringe with wavelength $λ_{2}$ .
$∴ n_{1} β_{1} = n_{2} β_{2}$
$\Rightarrow n_{1} \frac{λ_{1} D}{d} = n_{2} \frac{λ_{2} D}{d}$
$\Rightarrow n_{1} λ_{1} = n_{2} λ_{2}$ $(1)$
Also at first position, the $n^{th}$ bright fringe of one will coincide with $(n + 1)^{th}$ bright of other.
If $λ_{2} < λ_{1}$ ,
So then $n_{2} > n_{1}$
$\Rightarrow n_{2} = n_{1} + 1$ $(2)$
Using equation (2) and equation (1),
$n_{1} λ_{1} = (n_{1} + 1) λ_{2}$
$\Rightarrow (n_{1}) (650 \times 10^{- 9}) = (n_{1} + 1) (520 \times 10^{- 9})$
$\Rightarrow 65 n_{1} = 52 n_{1} + 52$
$\Rightarrow 13 n_{1} = 52$
$\Rightarrow n_{1} = 4$
Thus, $y = n_{1} β_{1}$
$= n_{1} \frac{λ_{1} D}{d}$
$= 4 [\frac{(6.5 \times 10^{- 7}) (1.5)}{3 \times 10^{- 3}}]$
$= 1.3 \times 10^{- 3} m$
$= 1.3 m m$

PHXII10:WAVE OPTICS

368133 The source that illuminates the double - slit in 'double - slit interference experiment' emits two distinct monochromatic waves of wavelength $500 n m$ and $600 n m$ , each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one move out of this central region, the two fringe systems are gradually out of step such that maximum due to one wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $n m$ is:

1 2000

2 3000

3 1000

4 1500

Explanation:

Let $λ_{1} = 500 n m, λ_{2} = 600 n m$
When maxima of first one coincides with minima of second one
$\frac{n λ_{1} D}{d} = \frac{(2 m + 1) λ_{2} D}{2 d}$
$\frac{n}{2 m + 1} = \frac{λ_{2}}{2 λ_{1}} = \frac{600}{2 \times 500} = \frac{3}{5}$
$n = \frac{3}{5} (2 m + 1)$
when $m = 2 \Rightarrow n = 3$
Path difference $P . d$ $= n λ_{1} = 1500 n m$

JEE - 2013

PHXII10:WAVE OPTICS

368134 In a Young’s double slit experiment, slits are separated by $0.5 m m,$ and the screen is placed $150 c m$ away. A beam of light consisting of two wavelengths, $650 n m a n d 520 n m,$ is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

1

9.75 m m

2

15.6 m m

3

1.56 m m

4

7.8 m m

Explanation:

For common maxima, $n_{1} λ_{1} = n_{2} λ_{2}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{520 \times 10^{- 9}}{650 \times 10^{- 9}} = \frac{4}{5}$
$F o r λ_{1}$
$y = \frac{n_{1} λ_{1} D}{d} = \frac{4 \times 650 \times 10^{- 9} \times 1.5}{0.5 \times 10^{- 3}} = 7.8 m m$

PHXII10:WAVE OPTICS

368135 Assertion :
In Young's double slit experiment using white light, the central fringe is white, the violet coloured fringe will be seen nearest to the central fringe.
Reason :
The interference patterns due to different component colours of white light overlap throughout the screen.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII10:WAVE OPTICS

368136 In a $Y D S E$ bi-chromatic light of wavelengths 400 $n m$ and 560 $n m$ are used. The distance between the slits is 0.1 $m m$ and the distance between the plane of the slits and the screen is 1 $m$ . Find the minimum distance between two successive regions of complete darkness

1

28 m m

2

32 m m

3

23 m m

4

42 m m

Explanation:

At the area of total darkness, in double slit apparatus, minima will occur for both the wavelengths which are incident simultaneously and normally.
$(\frac{2 n + 1}{2}) λ_{1} = \frac{(2 m + 1)}{2} λ_{2}$
$\Rightarrow \frac{2 n + 1}{2 m + 1} = \frac{λ_{2}}{λ_{1}}$
$\Rightarrow \frac{2 n + 1}{2 m + 1}$ $= \frac{560}{400} = \frac{7}{5} or 10 n = 14 m + 2$
By inspection, the two solutions are
(i) If $m_{1} = 2, n_{1} = 3$ (ii) If $m_{2} = 7, n_{2} = 10$ .
$∴$ Distance $Δ S = \frac{D λ_{1}}{d} [\frac{(2 n_{2} + 1) - (2 n_{1} + 1)}{2}]$
Now putting $n_{2} = 10$ and $n_{1} = 3$
$Δ S = 4 \times 7 \times 10^{- 3} m \Rightarrow Δ S = 28 m m$

PHXII10:WAVE OPTICS

368132 A beam of light consisting of two wavelengths, $650 n m$ and $520 n m$ is used to obtain interference fringes in a Young's double-slit experiment. What is the least distance from a central maximum where the bright fringes due to both the wavelengths coincide? The distance between the slits is $3 m m$ and the distance between the plane of the slits and screen is $150 c m$ .

1

3.4 m m

2

1.3 m m

3

5.4 m m

4

7.2 m m

Explanation:

Let $^{'} y^{'}$ be the linear distance between the centre of screen and the point at which the bright fringes due to both wavelength coincides. Let $n_{1}^{th}$ number of bright fringe with wavelength $λ_{1}$ coincides with $n_{2}^{th}$ number of bright fringe with wavelength $λ_{2}$ .
$∴ n_{1} β_{1} = n_{2} β_{2}$
$\Rightarrow n_{1} \frac{λ_{1} D}{d} = n_{2} \frac{λ_{2} D}{d}$
$\Rightarrow n_{1} λ_{1} = n_{2} λ_{2}$ $(1)$
Also at first position, the $n^{th}$ bright fringe of one will coincide with $(n + 1)^{th}$ bright of other.
If $λ_{2} < λ_{1}$ ,
So then $n_{2} > n_{1}$
$\Rightarrow n_{2} = n_{1} + 1$ $(2)$
Using equation (2) and equation (1),
$n_{1} λ_{1} = (n_{1} + 1) λ_{2}$
$\Rightarrow (n_{1}) (650 \times 10^{- 9}) = (n_{1} + 1) (520 \times 10^{- 9})$
$\Rightarrow 65 n_{1} = 52 n_{1} + 52$
$\Rightarrow 13 n_{1} = 52$
$\Rightarrow n_{1} = 4$
Thus, $y = n_{1} β_{1}$
$= n_{1} \frac{λ_{1} D}{d}$
$= 4 [\frac{(6.5 \times 10^{- 7}) (1.5)}{3 \times 10^{- 3}}]$
$= 1.3 \times 10^{- 3} m$
$= 1.3 m m$

PHXII10:WAVE OPTICS

368133 The source that illuminates the double - slit in 'double - slit interference experiment' emits two distinct monochromatic waves of wavelength $500 n m$ and $600 n m$ , each of them producing its own pattern on the screen. At the central point of the pattern when path difference is zero, maxima of both the patterns coincide and the resulting interference pattern is most distinct at the region of zero path difference. But as one move out of this central region, the two fringe systems are gradually out of step such that maximum due to one wavelength coincides with the minimum due to the other and the combined fringe system becomes completely indistinct. This may happen when path difference in $n m$ is:

1 2000

2 3000

3 1000

4 1500

Explanation:

Let $λ_{1} = 500 n m, λ_{2} = 600 n m$
When maxima of first one coincides with minima of second one
$\frac{n λ_{1} D}{d} = \frac{(2 m + 1) λ_{2} D}{2 d}$
$\frac{n}{2 m + 1} = \frac{λ_{2}}{2 λ_{1}} = \frac{600}{2 \times 500} = \frac{3}{5}$
$n = \frac{3}{5} (2 m + 1)$
when $m = 2 \Rightarrow n = 3$
Path difference $P . d$ $= n λ_{1} = 1500 n m$

JEE - 2013

PHXII10:WAVE OPTICS

368134 In a Young’s double slit experiment, slits are separated by $0.5 m m,$ and the screen is placed $150 c m$ away. A beam of light consisting of two wavelengths, $650 n m a n d 520 n m,$ is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is:

1

9.75 m m

2

15.6 m m

3

1.56 m m

4

7.8 m m

Explanation:

For common maxima, $n_{1} λ_{1} = n_{2} λ_{2}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} = \frac{520 \times 10^{- 9}}{650 \times 10^{- 9}} = \frac{4}{5}$
$F o r λ_{1}$
$y = \frac{n_{1} λ_{1} D}{d} = \frac{4 \times 650 \times 10^{- 9} \times 1.5}{0.5 \times 10^{- 3}} = 7.8 m m$

PHXII10:WAVE OPTICS

368135 Assertion :
In Young's double slit experiment using white light, the central fringe is white, the violet coloured fringe will be seen nearest to the central fringe.
Reason :
The interference patterns due to different component colours of white light overlap throughout the screen.

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

PHXII10:WAVE OPTICS

368136 In a $Y D S E$ bi-chromatic light of wavelengths 400 $n m$ and 560 $n m$ are used. The distance between the slits is 0.1 $m m$ and the distance between the plane of the slits and the screen is 1 $m$ . Find the minimum distance between two successive regions of complete darkness

1

28 m m

2

32 m m

3

23 m m

4

42 m m

Explanation:

At the area of total darkness, in double slit apparatus, minima will occur for both the wavelengths which are incident simultaneously and normally.
$(\frac{2 n + 1}{2}) λ_{1} = \frac{(2 m + 1)}{2} λ_{2}$
$\Rightarrow \frac{2 n + 1}{2 m + 1} = \frac{λ_{2}}{λ_{1}}$
$\Rightarrow \frac{2 n + 1}{2 m + 1}$ $= \frac{560}{400} = \frac{7}{5} or 10 n = 14 m + 2$
By inspection, the two solutions are
(i) If $m_{1} = 2, n_{1} = 3$ (ii) If $m_{2} = 7, n_{2} = 10$ .
$∴$ Distance $Δ S = \frac{D λ_{1}}{d} [\frac{(2 n_{2} + 1) - (2 n_{1} + 1)}{2}]$
Now putting $n_{2} = 10$ and $n_{1} = 3$
$Δ S = 4 \times 7 \times 10^{- 3} m \Rightarrow Δ S = 28 m m$

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation: