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PHXII10:WAVE OPTICS

368016 Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{t h}$ bright fringe in the medium lies where $5^{t h}$ dark fringe lies in air. The refractive index of the medium is nearly

1 1.59

2 1.69

3 1.78

4 1.25

Explanation:

${(y_{8})}_{B r i g h t, m e d i u m} = {(y_{5})}_{D a r k, a i r}$
$\frac{8 λ_{m} D}{d} = (\frac{2 (5) - 1}{2}) \frac{λ D}{d}$
$\frac{8 λ}{μ} \frac{D}{d} = \frac{9}{2} \frac{λ D}{d} \Rightarrow μ = \frac{16}{9} = 1.78$

NEET - 2017

PHXII10:WAVE OPTICS

368017 A light of wavelength $500 n m$ is incident on a Young's double slit. The distance between slit and screen is $D = 1.8 m$ and distance between slits is $d = 0.4 m m$ . If screen moves with a speed of $4 m s^{- 1}$ , then with what speed first maxima will move?

1

5 m m s^{- 1}

2

4 m m s^{- 1}

3

3 m m s^{- 1}

4

2 m m s^{- 1}

Explanation:

Given, wavelength of light,
$λ = 500 n m = 5 \times 10^{- 7} m$
Distance between slit and screen,
$D = 1.8 m$
Distance between two slits,
$d = 0.4 m m = 4 \times 10^{- 4} m$
velocity of screen $= \frac{d D}{d t} = 4 m s^{- 1}$
Speed of first maxima $= \frac{d β}{d t} =$ ?
In Young's double slit experiment, fringe width,
$β = \frac{D λ}{d}$
Differentiating with respect to $t$ , we get
$\frac{d β}{d t} = \frac{λ}{d} \cdot \frac{d D}{d t} = \frac{5 \times 10^{- 7}}{4 \times 10^{- 4}} \times 4$
$= 5 \times 10^{- 3} m s^{- 1} = 5 m m s^{- 1}$ .

AIIMS - 2019

PHXII10:WAVE OPTICS

368018 In Young’s double slit experiment, the slits are $3 m m$ apart. The wavelength of light used is $5000 \overset{\circ}{A}$ and the distance between the slits and the screen is $90 c m$ . The fringe width in $m m$ is

1 1.5

2 0.015

3 2

4 0.15

Explanation:

$β = \frac{λ D}{d} = \frac{5000 \times 10^{- 10} \times 0.9}{3 \times 10^{- 3}} m$
$= 0.15 m m$

PHXII10:WAVE OPTICS

368019 In Young’s double slit experiment, the central bright fringe can be identified

1 As it is wider than the other bright fringes

2 As it has greater intensity than the other bright fringes

3 By using white light instead of monochromatic light

4 As it is narrower than the other bright fringes

PHXII10:WAVE OPTICS

368016 Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{t h}$ bright fringe in the medium lies where $5^{t h}$ dark fringe lies in air. The refractive index of the medium is nearly

1 1.59

2 1.69

3 1.78

4 1.25

Explanation:

${(y_{8})}_{B r i g h t, m e d i u m} = {(y_{5})}_{D a r k, a i r}$
$\frac{8 λ_{m} D}{d} = (\frac{2 (5) - 1}{2}) \frac{λ D}{d}$
$\frac{8 λ}{μ} \frac{D}{d} = \frac{9}{2} \frac{λ D}{d} \Rightarrow μ = \frac{16}{9} = 1.78$

NEET - 2017

PHXII10:WAVE OPTICS

368017 A light of wavelength $500 n m$ is incident on a Young's double slit. The distance between slit and screen is $D = 1.8 m$ and distance between slits is $d = 0.4 m m$ . If screen moves with a speed of $4 m s^{- 1}$ , then with what speed first maxima will move?

1

5 m m s^{- 1}

2

4 m m s^{- 1}

3

3 m m s^{- 1}

4

2 m m s^{- 1}

Explanation:

Given, wavelength of light,
$λ = 500 n m = 5 \times 10^{- 7} m$
Distance between slit and screen,
$D = 1.8 m$
Distance between two slits,
$d = 0.4 m m = 4 \times 10^{- 4} m$
velocity of screen $= \frac{d D}{d t} = 4 m s^{- 1}$
Speed of first maxima $= \frac{d β}{d t} =$ ?
In Young's double slit experiment, fringe width,
$β = \frac{D λ}{d}$
Differentiating with respect to $t$ , we get
$\frac{d β}{d t} = \frac{λ}{d} \cdot \frac{d D}{d t} = \frac{5 \times 10^{- 7}}{4 \times 10^{- 4}} \times 4$
$= 5 \times 10^{- 3} m s^{- 1} = 5 m m s^{- 1}$ .

AIIMS - 2019

PHXII10:WAVE OPTICS

368018 In Young’s double slit experiment, the slits are $3 m m$ apart. The wavelength of light used is $5000 \overset{\circ}{A}$ and the distance between the slits and the screen is $90 c m$ . The fringe width in $m m$ is

1 1.5

2 0.015

3 2

4 0.15

Explanation:

$β = \frac{λ D}{d} = \frac{5000 \times 10^{- 10} \times 0.9}{3 \times 10^{- 3}} m$
$= 0.15 m m$

PHXII10:WAVE OPTICS

368019 In Young’s double slit experiment, the central bright fringe can be identified

1 As it is wider than the other bright fringes

2 As it has greater intensity than the other bright fringes

3 By using white light instead of monochromatic light

4 As it is narrower than the other bright fringes

PHXII10:WAVE OPTICS

368016 Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{t h}$ bright fringe in the medium lies where $5^{t h}$ dark fringe lies in air. The refractive index of the medium is nearly

1 1.59

2 1.69

3 1.78

4 1.25

Explanation:

${(y_{8})}_{B r i g h t, m e d i u m} = {(y_{5})}_{D a r k, a i r}$
$\frac{8 λ_{m} D}{d} = (\frac{2 (5) - 1}{2}) \frac{λ D}{d}$
$\frac{8 λ}{μ} \frac{D}{d} = \frac{9}{2} \frac{λ D}{d} \Rightarrow μ = \frac{16}{9} = 1.78$

NEET - 2017

PHXII10:WAVE OPTICS

368017 A light of wavelength $500 n m$ is incident on a Young's double slit. The distance between slit and screen is $D = 1.8 m$ and distance between slits is $d = 0.4 m m$ . If screen moves with a speed of $4 m s^{- 1}$ , then with what speed first maxima will move?

1

5 m m s^{- 1}

2

4 m m s^{- 1}

3

3 m m s^{- 1}

4

2 m m s^{- 1}

Explanation:

Given, wavelength of light,
$λ = 500 n m = 5 \times 10^{- 7} m$
Distance between slit and screen,
$D = 1.8 m$
Distance between two slits,
$d = 0.4 m m = 4 \times 10^{- 4} m$
velocity of screen $= \frac{d D}{d t} = 4 m s^{- 1}$
Speed of first maxima $= \frac{d β}{d t} =$ ?
In Young's double slit experiment, fringe width,
$β = \frac{D λ}{d}$
Differentiating with respect to $t$ , we get
$\frac{d β}{d t} = \frac{λ}{d} \cdot \frac{d D}{d t} = \frac{5 \times 10^{- 7}}{4 \times 10^{- 4}} \times 4$
$= 5 \times 10^{- 3} m s^{- 1} = 5 m m s^{- 1}$ .

AIIMS - 2019

PHXII10:WAVE OPTICS

368018 In Young’s double slit experiment, the slits are $3 m m$ apart. The wavelength of light used is $5000 \overset{\circ}{A}$ and the distance between the slits and the screen is $90 c m$ . The fringe width in $m m$ is

1 1.5

2 0.015

3 2

4 0.15

Explanation:

$β = \frac{λ D}{d} = \frac{5000 \times 10^{- 10} \times 0.9}{3 \times 10^{- 3}} m$
$= 0.15 m m$

PHXII10:WAVE OPTICS

368019 In Young’s double slit experiment, the central bright fringe can be identified

1 As it is wider than the other bright fringes

2 As it has greater intensity than the other bright fringes

3 By using white light instead of monochromatic light

4 As it is narrower than the other bright fringes

PHXII10:WAVE OPTICS

368016 Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that $8^{t h}$ bright fringe in the medium lies where $5^{t h}$ dark fringe lies in air. The refractive index of the medium is nearly

1 1.59

2 1.69

3 1.78

4 1.25

Explanation:

${(y_{8})}_{B r i g h t, m e d i u m} = {(y_{5})}_{D a r k, a i r}$
$\frac{8 λ_{m} D}{d} = (\frac{2 (5) - 1}{2}) \frac{λ D}{d}$
$\frac{8 λ}{μ} \frac{D}{d} = \frac{9}{2} \frac{λ D}{d} \Rightarrow μ = \frac{16}{9} = 1.78$

NEET - 2017

PHXII10:WAVE OPTICS

368017 A light of wavelength $500 n m$ is incident on a Young's double slit. The distance between slit and screen is $D = 1.8 m$ and distance between slits is $d = 0.4 m m$ . If screen moves with a speed of $4 m s^{- 1}$ , then with what speed first maxima will move?

1

5 m m s^{- 1}

2

4 m m s^{- 1}

3

3 m m s^{- 1}

4

2 m m s^{- 1}

Explanation:

Given, wavelength of light,
$λ = 500 n m = 5 \times 10^{- 7} m$
Distance between slit and screen,
$D = 1.8 m$
Distance between two slits,
$d = 0.4 m m = 4 \times 10^{- 4} m$
velocity of screen $= \frac{d D}{d t} = 4 m s^{- 1}$
Speed of first maxima $= \frac{d β}{d t} =$ ?
In Young's double slit experiment, fringe width,
$β = \frac{D λ}{d}$
Differentiating with respect to $t$ , we get
$\frac{d β}{d t} = \frac{λ}{d} \cdot \frac{d D}{d t} = \frac{5 \times 10^{- 7}}{4 \times 10^{- 4}} \times 4$
$= 5 \times 10^{- 3} m s^{- 1} = 5 m m s^{- 1}$ .

AIIMS - 2019

PHXII10:WAVE OPTICS

368018 In Young’s double slit experiment, the slits are $3 m m$ apart. The wavelength of light used is $5000 \overset{\circ}{A}$ and the distance between the slits and the screen is $90 c m$ . The fringe width in $m m$ is

1 1.5

2 0.015

3 2

4 0.15

Explanation:

$β = \frac{λ D}{d} = \frac{5000 \times 10^{- 10} \times 0.9}{3 \times 10^{- 3}} m$
$= 0.15 m m$

PHXII10:WAVE OPTICS

368019 In Young’s double slit experiment, the central bright fringe can be identified

1 As it is wider than the other bright fringes

2 As it has greater intensity than the other bright fringes

3 By using white light instead of monochromatic light

4 As it is narrower than the other bright fringes

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