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PHXII10:WAVE OPTICS

367972 A young’s double slit experiment uses a monochromatic rectangular source. The shape of the interference fringes formed on a screen is

1 Hyperbola

2 Circle

3 Straight line

4 Parabola

PHXII10:WAVE OPTICS

367973 Fringes are produced by a Fresnel biprism in the focal plane of an eye-piece which is at a distance of $1 m$ from the centre slit. A lens inserted between the biprism and eye-piece gives two images of the slit in two positions. In first case, the images of the slit are $9 \times 10^{- 3} m$ and in the second position the images are $4 \times 10^{- 3} m$ apart. If the sodium light $(λ = 6000 \times 10^{- 10} m)$ is used, then the distance between the interference fringes is found to be $1 \times 10^{- M} m$ . What is the value of $M$ ? (Take $π = 3.14$ )

1 1

2 5

3 7

4 4

Explanation:

When the distance between the screen and the object is greater than four times the focal length of a convex lens, there are two positions of the lens where it forms a real image of the object on the screen. If $d_{1}$ and $d_{2}$ are the sizes of the images in the two positions then $d$ (size of the object) $= \sqrt{d_{1} d_{2}}$
$∴ d$ (distance between virtual sources)
$= \sqrt{9 \times 10^{- 3} \times 4 \times 10^{- 3}}$
But, $β = \frac{λ D}{d}$
$∴ β = \frac{6000 \times 10^{- 10} \times 1}{\sqrt{4 \times 9 \times 10^{- 6}}}$
$= \frac{6000 \times 10^{- 10}}{6 \times 10^{- 3}} = 1 \times 10^{- 4}$
$∴ M = 4$

PHXII10:WAVE OPTICS

367974 Two coherent sources have initial phase difference zero. A circular screen surrounds the two sources as shown in the figure. Find the total number of maxima’s formed on the screen $(R >> d, λ = 3 d)$
supporting img

1 14

2 7

3 12

4 5

Explanation:

The condition for maxima is $d \sin θ = n λ$
$3 λ \sin θ = n λ \Rightarrow \sin θ = \frac{n}{3}$
The possible values of $n$ are $0, 1, 2, 3.$
The total possible maxima's on the circular screen are shown in the figure.

supporting img

The total maxima's on the screen are equal to 12.

PHXII10:WAVE OPTICS

367975 In Young’s double - slit experiment, how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe $(λ = 2000 \overset{\circ}{A}, d = 7000 \overset{\circ}{A})$ ?

1 7

2 12

3 4

4 18

Explanation:

For maximum intensity on the screen,
$d \sin θ = n λ$
or $\sin θ = \frac{n λ}{d} = \frac{(n) (2000)}{(7000)} = \frac{n}{3.5}$
$\sin θ \geq 1 \Rightarrow n = 0, 1, 2, 3$
Thus, only seven maxima can be obtained on both sides of the screen.

PHXII10:WAVE OPTICS

367972 A young’s double slit experiment uses a monochromatic rectangular source. The shape of the interference fringes formed on a screen is

1 Hyperbola

2 Circle

3 Straight line

4 Parabola

PHXII10:WAVE OPTICS

367973 Fringes are produced by a Fresnel biprism in the focal plane of an eye-piece which is at a distance of $1 m$ from the centre slit. A lens inserted between the biprism and eye-piece gives two images of the slit in two positions. In first case, the images of the slit are $9 \times 10^{- 3} m$ and in the second position the images are $4 \times 10^{- 3} m$ apart. If the sodium light $(λ = 6000 \times 10^{- 10} m)$ is used, then the distance between the interference fringes is found to be $1 \times 10^{- M} m$ . What is the value of $M$ ? (Take $π = 3.14$ )

1 1

2 5

3 7

4 4

Explanation:

When the distance between the screen and the object is greater than four times the focal length of a convex lens, there are two positions of the lens where it forms a real image of the object on the screen. If $d_{1}$ and $d_{2}$ are the sizes of the images in the two positions then $d$ (size of the object) $= \sqrt{d_{1} d_{2}}$
$∴ d$ (distance between virtual sources)
$= \sqrt{9 \times 10^{- 3} \times 4 \times 10^{- 3}}$
But, $β = \frac{λ D}{d}$
$∴ β = \frac{6000 \times 10^{- 10} \times 1}{\sqrt{4 \times 9 \times 10^{- 6}}}$
$= \frac{6000 \times 10^{- 10}}{6 \times 10^{- 3}} = 1 \times 10^{- 4}$
$∴ M = 4$

PHXII10:WAVE OPTICS

367974 Two coherent sources have initial phase difference zero. A circular screen surrounds the two sources as shown in the figure. Find the total number of maxima’s formed on the screen $(R >> d, λ = 3 d)$
supporting img

1 14

2 7

3 12

4 5

Explanation:

The condition for maxima is $d \sin θ = n λ$
$3 λ \sin θ = n λ \Rightarrow \sin θ = \frac{n}{3}$
The possible values of $n$ are $0, 1, 2, 3.$
The total possible maxima's on the circular screen are shown in the figure.

supporting img

The total maxima's on the screen are equal to 12.

PHXII10:WAVE OPTICS

367975 In Young’s double - slit experiment, how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe $(λ = 2000 \overset{\circ}{A}, d = 7000 \overset{\circ}{A})$ ?

1 7

2 12

3 4

4 18

Explanation:

For maximum intensity on the screen,
$d \sin θ = n λ$
or $\sin θ = \frac{n λ}{d} = \frac{(n) (2000)}{(7000)} = \frac{n}{3.5}$
$\sin θ \geq 1 \Rightarrow n = 0, 1, 2, 3$
Thus, only seven maxima can be obtained on both sides of the screen.

PHXII10:WAVE OPTICS

367972 A young’s double slit experiment uses a monochromatic rectangular source. The shape of the interference fringes formed on a screen is

1 Hyperbola

2 Circle

3 Straight line

4 Parabola

PHXII10:WAVE OPTICS

367973 Fringes are produced by a Fresnel biprism in the focal plane of an eye-piece which is at a distance of $1 m$ from the centre slit. A lens inserted between the biprism and eye-piece gives two images of the slit in two positions. In first case, the images of the slit are $9 \times 10^{- 3} m$ and in the second position the images are $4 \times 10^{- 3} m$ apart. If the sodium light $(λ = 6000 \times 10^{- 10} m)$ is used, then the distance between the interference fringes is found to be $1 \times 10^{- M} m$ . What is the value of $M$ ? (Take $π = 3.14$ )

1 1

2 5

3 7

4 4

Explanation:

When the distance between the screen and the object is greater than four times the focal length of a convex lens, there are two positions of the lens where it forms a real image of the object on the screen. If $d_{1}$ and $d_{2}$ are the sizes of the images in the two positions then $d$ (size of the object) $= \sqrt{d_{1} d_{2}}$
$∴ d$ (distance between virtual sources)
$= \sqrt{9 \times 10^{- 3} \times 4 \times 10^{- 3}}$
But, $β = \frac{λ D}{d}$
$∴ β = \frac{6000 \times 10^{- 10} \times 1}{\sqrt{4 \times 9 \times 10^{- 6}}}$
$= \frac{6000 \times 10^{- 10}}{6 \times 10^{- 3}} = 1 \times 10^{- 4}$
$∴ M = 4$

PHXII10:WAVE OPTICS

367974 Two coherent sources have initial phase difference zero. A circular screen surrounds the two sources as shown in the figure. Find the total number of maxima’s formed on the screen $(R >> d, λ = 3 d)$
supporting img

1 14

2 7

3 12

4 5

Explanation:

The condition for maxima is $d \sin θ = n λ$
$3 λ \sin θ = n λ \Rightarrow \sin θ = \frac{n}{3}$
The possible values of $n$ are $0, 1, 2, 3.$
The total possible maxima's on the circular screen are shown in the figure.

supporting img

The total maxima's on the screen are equal to 12.

PHXII10:WAVE OPTICS

367975 In Young’s double - slit experiment, how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe $(λ = 2000 \overset{\circ}{A}, d = 7000 \overset{\circ}{A})$ ?

1 7

2 12

3 4

4 18

Explanation:

For maximum intensity on the screen,
$d \sin θ = n λ$
or $\sin θ = \frac{n λ}{d} = \frac{(n) (2000)}{(7000)} = \frac{n}{3.5}$
$\sin θ \geq 1 \Rightarrow n = 0, 1, 2, 3$
Thus, only seven maxima can be obtained on both sides of the screen.

PHXII10:WAVE OPTICS

367972 A young’s double slit experiment uses a monochromatic rectangular source. The shape of the interference fringes formed on a screen is

1 Hyperbola

2 Circle

3 Straight line

4 Parabola

PHXII10:WAVE OPTICS

367973 Fringes are produced by a Fresnel biprism in the focal plane of an eye-piece which is at a distance of $1 m$ from the centre slit. A lens inserted between the biprism and eye-piece gives two images of the slit in two positions. In first case, the images of the slit are $9 \times 10^{- 3} m$ and in the second position the images are $4 \times 10^{- 3} m$ apart. If the sodium light $(λ = 6000 \times 10^{- 10} m)$ is used, then the distance between the interference fringes is found to be $1 \times 10^{- M} m$ . What is the value of $M$ ? (Take $π = 3.14$ )

1 1

2 5

3 7

4 4

Explanation:

When the distance between the screen and the object is greater than four times the focal length of a convex lens, there are two positions of the lens where it forms a real image of the object on the screen. If $d_{1}$ and $d_{2}$ are the sizes of the images in the two positions then $d$ (size of the object) $= \sqrt{d_{1} d_{2}}$
$∴ d$ (distance between virtual sources)
$= \sqrt{9 \times 10^{- 3} \times 4 \times 10^{- 3}}$
But, $β = \frac{λ D}{d}$
$∴ β = \frac{6000 \times 10^{- 10} \times 1}{\sqrt{4 \times 9 \times 10^{- 6}}}$
$= \frac{6000 \times 10^{- 10}}{6 \times 10^{- 3}} = 1 \times 10^{- 4}$
$∴ M = 4$

PHXII10:WAVE OPTICS

367974 Two coherent sources have initial phase difference zero. A circular screen surrounds the two sources as shown in the figure. Find the total number of maxima’s formed on the screen $(R >> d, λ = 3 d)$
supporting img

1 14

2 7

3 12

4 5

Explanation:

The condition for maxima is $d \sin θ = n λ$
$3 λ \sin θ = n λ \Rightarrow \sin θ = \frac{n}{3}$
The possible values of $n$ are $0, 1, 2, 3.$
The total possible maxima's on the circular screen are shown in the figure.

supporting img

The total maxima's on the screen are equal to 12.

PHXII10:WAVE OPTICS

367975 In Young’s double - slit experiment, how many maximas can be obtained on a screen (including the central maximum) on both sides of the central fringe $(λ = 2000 \overset{\circ}{A}, d = 7000 \overset{\circ}{A})$ ?

1 7

2 12

3 4

4 18

Explanation:

For maximum intensity on the screen,
$d \sin θ = n λ$
or $\sin θ = \frac{n λ}{d} = \frac{(n) (2000)}{(7000)} = \frac{n}{3.5}$
$\sin θ \geq 1 \Rightarrow n = 0, 1, 2, 3$
Thus, only seven maxima can be obtained on both sides of the screen.