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PHXII10:WAVE OPTICS

367700 The diffraction pattern of a light of wavelength $400 n m$ diffracting from a slit of width $0.2 m m$ is focused on the focal plane of a convex lens of focal length $100 c m .$ The width of the $1^{st}$ secondary maxima will be

1

0.2 m m

2

2 c m

3

2 m m

4

0.02 m m

Explanation:

For diffraction, $d \sin θ = n λ$
$\Rightarrow \sin θ = \frac{n λ}{d}$
where $θ$ is diffraction angle.
For the width of $1^{st}$ secondary maxima, the diffraction angle of first and second dark fringes are required, we have
$θ_{1} \approx \sin θ_{1} = \frac{1 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.002$ (using small angle approximation)
and $θ_{2} \approx \sin θ_{2} = \frac{2 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.004$
$\Rightarrow θ_{1} = 0.002 r a d, θ_{2} = 0.004 r a d$
Now, $y = D \tan θ$ , where $y$ is the position of the fringe, $D$ is distance from the screen.
$∴ y_{1} = 100 \times \tan θ_{1} = 100 \times 0.002 = 0.2 c m$
$y_{2} = 100 \times \tan θ_{2} = 100 \times 0.004 = 0.4 c m$
$∴$ Width of secondary maximum,
$β = y_{2} - y_{1} = 0.4 - 0.2 = 0.2 c m$
$β = 2 m m$

JEE - 2024

PHXII10:WAVE OPTICS

367701 In a single slit diffraction, the width of slit is $0.5 c m,$ focal length of lens is $40 c m$ , and wavelength of light is $4890 \overset{\circ}{A}$ The distance of first dark fringe is

1

2 \times 10^{- 5} m

2

4 \times 10^{- 5} m

3

6 \times 10^{- 5} m

4

8 \times 10^{- 5} m

Explanation:

$y = \frac{λ F}{a}$
$y = 4890 \times 10^{- 10} \times 80 = 3.912 \times 10^{- 5} \approx 4 \times 10^{- 5} m$

PHXII10:WAVE OPTICS

367702 A slit $4 c m$ wide is irradiated with microwaves of wavelength $2 c m$ . What will be angular spread of central maximum, assuming incidence normal to the plane of the slit?

1

30^{\circ}

2

40^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

$a = 4 cm = 4 \times 10^{- 2} m$
$λ = 2 cm = 2 \times 10^{- 2} m$
Angular spread of central maximum ( $2 θ)$ is $a \sin θ = λ n$
$\sin θ = \frac{2 \times 10^{- 2}}{4 \times 10^{- 2}} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = 30^{\circ}$
Angular speed $= 2 θ = 2 \times 30^{\circ} = 60^{\circ}$ .

PHXII10:WAVE OPTICS

367703 The slit width, when a light of wavelength $6500 \overset{\circ}{A}$ is incident on a slit, if first minima for red light is at $30^{\circ}$ , is

1

1 \times 10^{- 6} m

2

5.2 \times 10^{- 6} m

3

1.3 \times 10^{- 6} m

4

2.6 \times 10^{- 6} m

Explanation:

When rays of monochromatic light of wavelength $λ$ are incident on a diffraction grating in which slit separation is $d$ , then for angle of diffraction $θ$ , the following relation holds true.
$d \sin θ = n λ$
where, $n$ is called the spectrum order.
Given, $n = 1, λ = 6500 \overset{\circ}{A}$
$= 6500 \times 10^{- 10} m$
and $\sin 30^{\circ} = \frac{1}{2}$
$\Rightarrow d = \frac{n λ}{\sin 30^{\circ}} = \frac{6500 \times 10^{- 10}}{(1 / 2)}$
$\Rightarrow d = 1.3 \times 10^{- 6} m$

PHXII10:WAVE OPTICS

367700 The diffraction pattern of a light of wavelength $400 n m$ diffracting from a slit of width $0.2 m m$ is focused on the focal plane of a convex lens of focal length $100 c m .$ The width of the $1^{st}$ secondary maxima will be

1

0.2 m m

2

2 c m

3

2 m m

4

0.02 m m

Explanation:

For diffraction, $d \sin θ = n λ$
$\Rightarrow \sin θ = \frac{n λ}{d}$
where $θ$ is diffraction angle.
For the width of $1^{st}$ secondary maxima, the diffraction angle of first and second dark fringes are required, we have
$θ_{1} \approx \sin θ_{1} = \frac{1 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.002$ (using small angle approximation)
and $θ_{2} \approx \sin θ_{2} = \frac{2 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.004$
$\Rightarrow θ_{1} = 0.002 r a d, θ_{2} = 0.004 r a d$
Now, $y = D \tan θ$ , where $y$ is the position of the fringe, $D$ is distance from the screen.
$∴ y_{1} = 100 \times \tan θ_{1} = 100 \times 0.002 = 0.2 c m$
$y_{2} = 100 \times \tan θ_{2} = 100 \times 0.004 = 0.4 c m$
$∴$ Width of secondary maximum,
$β = y_{2} - y_{1} = 0.4 - 0.2 = 0.2 c m$
$β = 2 m m$

JEE - 2024

PHXII10:WAVE OPTICS

367701 In a single slit diffraction, the width of slit is $0.5 c m,$ focal length of lens is $40 c m$ , and wavelength of light is $4890 \overset{\circ}{A}$ The distance of first dark fringe is

1

2 \times 10^{- 5} m

2

4 \times 10^{- 5} m

3

6 \times 10^{- 5} m

4

8 \times 10^{- 5} m

Explanation:

$y = \frac{λ F}{a}$
$y = 4890 \times 10^{- 10} \times 80 = 3.912 \times 10^{- 5} \approx 4 \times 10^{- 5} m$

PHXII10:WAVE OPTICS

367702 A slit $4 c m$ wide is irradiated with microwaves of wavelength $2 c m$ . What will be angular spread of central maximum, assuming incidence normal to the plane of the slit?

1

30^{\circ}

2

40^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

$a = 4 cm = 4 \times 10^{- 2} m$
$λ = 2 cm = 2 \times 10^{- 2} m$
Angular spread of central maximum ( $2 θ)$ is $a \sin θ = λ n$
$\sin θ = \frac{2 \times 10^{- 2}}{4 \times 10^{- 2}} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = 30^{\circ}$
Angular speed $= 2 θ = 2 \times 30^{\circ} = 60^{\circ}$ .

PHXII10:WAVE OPTICS

367703 The slit width, when a light of wavelength $6500 \overset{\circ}{A}$ is incident on a slit, if first minima for red light is at $30^{\circ}$ , is

1

1 \times 10^{- 6} m

2

5.2 \times 10^{- 6} m

3

1.3 \times 10^{- 6} m

4

2.6 \times 10^{- 6} m

Explanation:

When rays of monochromatic light of wavelength $λ$ are incident on a diffraction grating in which slit separation is $d$ , then for angle of diffraction $θ$ , the following relation holds true.
$d \sin θ = n λ$
where, $n$ is called the spectrum order.
Given, $n = 1, λ = 6500 \overset{\circ}{A}$
$= 6500 \times 10^{- 10} m$
and $\sin 30^{\circ} = \frac{1}{2}$
$\Rightarrow d = \frac{n λ}{\sin 30^{\circ}} = \frac{6500 \times 10^{- 10}}{(1 / 2)}$
$\Rightarrow d = 1.3 \times 10^{- 6} m$

PHXII10:WAVE OPTICS

367700 The diffraction pattern of a light of wavelength $400 n m$ diffracting from a slit of width $0.2 m m$ is focused on the focal plane of a convex lens of focal length $100 c m .$ The width of the $1^{st}$ secondary maxima will be

1

0.2 m m

2

2 c m

3

2 m m

4

0.02 m m

Explanation:

For diffraction, $d \sin θ = n λ$
$\Rightarrow \sin θ = \frac{n λ}{d}$
where $θ$ is diffraction angle.
For the width of $1^{st}$ secondary maxima, the diffraction angle of first and second dark fringes are required, we have
$θ_{1} \approx \sin θ_{1} = \frac{1 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.002$ (using small angle approximation)
and $θ_{2} \approx \sin θ_{2} = \frac{2 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.004$
$\Rightarrow θ_{1} = 0.002 r a d, θ_{2} = 0.004 r a d$
Now, $y = D \tan θ$ , where $y$ is the position of the fringe, $D$ is distance from the screen.
$∴ y_{1} = 100 \times \tan θ_{1} = 100 \times 0.002 = 0.2 c m$
$y_{2} = 100 \times \tan θ_{2} = 100 \times 0.004 = 0.4 c m$
$∴$ Width of secondary maximum,
$β = y_{2} - y_{1} = 0.4 - 0.2 = 0.2 c m$
$β = 2 m m$

JEE - 2024

PHXII10:WAVE OPTICS

367701 In a single slit diffraction, the width of slit is $0.5 c m,$ focal length of lens is $40 c m$ , and wavelength of light is $4890 \overset{\circ}{A}$ The distance of first dark fringe is

1

2 \times 10^{- 5} m

2

4 \times 10^{- 5} m

3

6 \times 10^{- 5} m

4

8 \times 10^{- 5} m

Explanation:

$y = \frac{λ F}{a}$
$y = 4890 \times 10^{- 10} \times 80 = 3.912 \times 10^{- 5} \approx 4 \times 10^{- 5} m$

PHXII10:WAVE OPTICS

367702 A slit $4 c m$ wide is irradiated with microwaves of wavelength $2 c m$ . What will be angular spread of central maximum, assuming incidence normal to the plane of the slit?

1

30^{\circ}

2

40^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

$a = 4 cm = 4 \times 10^{- 2} m$
$λ = 2 cm = 2 \times 10^{- 2} m$
Angular spread of central maximum ( $2 θ)$ is $a \sin θ = λ n$
$\sin θ = \frac{2 \times 10^{- 2}}{4 \times 10^{- 2}} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = 30^{\circ}$
Angular speed $= 2 θ = 2 \times 30^{\circ} = 60^{\circ}$ .

PHXII10:WAVE OPTICS

367703 The slit width, when a light of wavelength $6500 \overset{\circ}{A}$ is incident on a slit, if first minima for red light is at $30^{\circ}$ , is

1

1 \times 10^{- 6} m

2

5.2 \times 10^{- 6} m

3

1.3 \times 10^{- 6} m

4

2.6 \times 10^{- 6} m

Explanation:

When rays of monochromatic light of wavelength $λ$ are incident on a diffraction grating in which slit separation is $d$ , then for angle of diffraction $θ$ , the following relation holds true.
$d \sin θ = n λ$
where, $n$ is called the spectrum order.
Given, $n = 1, λ = 6500 \overset{\circ}{A}$
$= 6500 \times 10^{- 10} m$
and $\sin 30^{\circ} = \frac{1}{2}$
$\Rightarrow d = \frac{n λ}{\sin 30^{\circ}} = \frac{6500 \times 10^{- 10}}{(1 / 2)}$
$\Rightarrow d = 1.3 \times 10^{- 6} m$

PHXII10:WAVE OPTICS

367700 The diffraction pattern of a light of wavelength $400 n m$ diffracting from a slit of width $0.2 m m$ is focused on the focal plane of a convex lens of focal length $100 c m .$ The width of the $1^{st}$ secondary maxima will be

1

0.2 m m

2

2 c m

3

2 m m

4

0.02 m m

Explanation:

For diffraction, $d \sin θ = n λ$
$\Rightarrow \sin θ = \frac{n λ}{d}$
where $θ$ is diffraction angle.
For the width of $1^{st}$ secondary maxima, the diffraction angle of first and second dark fringes are required, we have
$θ_{1} \approx \sin θ_{1} = \frac{1 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.002$ (using small angle approximation)
and $θ_{2} \approx \sin θ_{2} = \frac{2 \times 400 \times 10^{- 9}}{0.2 \times 10^{- 3}} = 0.004$
$\Rightarrow θ_{1} = 0.002 r a d, θ_{2} = 0.004 r a d$
Now, $y = D \tan θ$ , where $y$ is the position of the fringe, $D$ is distance from the screen.
$∴ y_{1} = 100 \times \tan θ_{1} = 100 \times 0.002 = 0.2 c m$
$y_{2} = 100 \times \tan θ_{2} = 100 \times 0.004 = 0.4 c m$
$∴$ Width of secondary maximum,
$β = y_{2} - y_{1} = 0.4 - 0.2 = 0.2 c m$
$β = 2 m m$

JEE - 2024

PHXII10:WAVE OPTICS

367701 In a single slit diffraction, the width of slit is $0.5 c m,$ focal length of lens is $40 c m$ , and wavelength of light is $4890 \overset{\circ}{A}$ The distance of first dark fringe is

1

2 \times 10^{- 5} m

2

4 \times 10^{- 5} m

3

6 \times 10^{- 5} m

4

8 \times 10^{- 5} m

Explanation:

$y = \frac{λ F}{a}$
$y = 4890 \times 10^{- 10} \times 80 = 3.912 \times 10^{- 5} \approx 4 \times 10^{- 5} m$

PHXII10:WAVE OPTICS

367702 A slit $4 c m$ wide is irradiated with microwaves of wavelength $2 c m$ . What will be angular spread of central maximum, assuming incidence normal to the plane of the slit?

1

30^{\circ}

2

40^{\circ}

3

45^{\circ}

4

60^{\circ}

Explanation:

$a = 4 cm = 4 \times 10^{- 2} m$
$λ = 2 cm = 2 \times 10^{- 2} m$
Angular spread of central maximum ( $2 θ)$ is $a \sin θ = λ n$
$\sin θ = \frac{2 \times 10^{- 2}}{4 \times 10^{- 2}} \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = 30^{\circ}$
Angular speed $= 2 θ = 2 \times 30^{\circ} = 60^{\circ}$ .

PHXII10:WAVE OPTICS

367703 The slit width, when a light of wavelength $6500 \overset{\circ}{A}$ is incident on a slit, if first minima for red light is at $30^{\circ}$ , is

1

1 \times 10^{- 6} m

2

5.2 \times 10^{- 6} m

3

1.3 \times 10^{- 6} m

4

2.6 \times 10^{- 6} m

Explanation:

When rays of monochromatic light of wavelength $λ$ are incident on a diffraction grating in which slit separation is $d$ , then for angle of diffraction $θ$ , the following relation holds true.
$d \sin θ = n λ$
where, $n$ is called the spectrum order.
Given, $n = 1, λ = 6500 \overset{\circ}{A}$
$= 6500 \times 10^{- 10} m$
and $\sin 30^{\circ} = \frac{1}{2}$
$\Rightarrow d = \frac{n λ}{\sin 30^{\circ}} = \frac{6500 \times 10^{- 10}}{(1 / 2)}$
$\Rightarrow d = 1.3 \times 10^{- 6} m$

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