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PHXII10:WAVE OPTICS

367726 Angular width of central maxima is $π / 2$ when a slit of width $^{'} A^{'}$ is illuminated by a light of wavelength $7000 \overset{\circ}{A}$ then $a =$

1

9 \times 10^{- 9} m

2

8.9 \times 10^{- 7} m

3

9 \times 10^{- 7} m

4

9.8 \times 10^{- 7} m

Explanation:

$\sin θ = \frac{λ}{a}, θ = 45^{\circ}, λ = 7000 \overset{o}{A}$
$a = \sqrt{2} λ = 9.8 \times 10^{- 7} m$

PHXII10:WAVE OPTICS

367727 A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $λ$ . If the second and fourth minima in the difdfraction pattern at distance $1 m$ from the slit are at $3 c m$ and $6 c m$ respectively from the central maximum what is the width of the central maximum? (i.e., distance between first minimum on either side of the central maximum)

1

4.5 c m

2

1.5 c m

3

6.0 c m

4

3.0 c m

Explanation:

The position of $n^{t h}$ minima is
$y = \frac{n λ D}{a}, y_{2} = \frac{2 λ D}{a}, y_{4} = \frac{4 λ D}{a}$
$y_{4} - y_{2} = \frac{2 λ D}{a} = 3 c m$
Width of central maxima is $\frac{2 λ D}{a} = 3 c m$

JEE - 2017

PHXII10:WAVE OPTICS

367728 In a double slit experiment, the two slits are 1 $m m$ apart and the screen is placed $1 m$ away. A monochromatic light wavelength $500 n m$ is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

1

0.1 m m

2

0.5 m m

3

0.02 m m

4

0.2 m m

Explanation:

Here, distance between two slits,
$d = 1 m m = 10^{- 3} m, D = 1 m$
$λ = 500 n m = 500 \times 10^{- 9} m$
width of central maxima in single slit diffraction pattern $\frac{2 λ D}{a}$
Fringe width in double slit experiment $β = \frac{λ D}{d}$
So, required condition $\frac{10 λ D}{d} = \frac{2 λ D}{a}$
$\Rightarrow a = \frac{d}{5} = \frac{1}{5} \times 10^{- 3} m = 0.2 m m$

PHXII10:WAVE OPTICS

367729 In a diffraction pattern due to a single slit of width $^{'} a^{'}$ , the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \overset{\circ}{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

1

\sin^{- 1} (\frac{1}{4})

2

\sin^{- 1} (\frac{2}{3})

3

\sin^{- 1} (\frac{1}{2})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

For first minima, $\sin 30^{\circ} = \frac{λ}{a} = \frac{1}{2}$
First secondary maxima will be at
$\sin θ = \frac{3 λ}{2 a} = \frac{3}{2} (\frac{1}{2}) \Rightarrow θ = \sin^{- 1} (\frac{3}{4})$

NEET - 2016

PHXII10:WAVE OPTICS

367726 Angular width of central maxima is $π / 2$ when a slit of width $^{'} A^{'}$ is illuminated by a light of wavelength $7000 \overset{\circ}{A}$ then $a =$

1

9 \times 10^{- 9} m

2

8.9 \times 10^{- 7} m

3

9 \times 10^{- 7} m

4

9.8 \times 10^{- 7} m

Explanation:

$\sin θ = \frac{λ}{a}, θ = 45^{\circ}, λ = 7000 \overset{o}{A}$
$a = \sqrt{2} λ = 9.8 \times 10^{- 7} m$

PHXII10:WAVE OPTICS

367727 A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $λ$ . If the second and fourth minima in the difdfraction pattern at distance $1 m$ from the slit are at $3 c m$ and $6 c m$ respectively from the central maximum what is the width of the central maximum? (i.e., distance between first minimum on either side of the central maximum)

1

4.5 c m

2

1.5 c m

3

6.0 c m

4

3.0 c m

Explanation:

The position of $n^{t h}$ minima is
$y = \frac{n λ D}{a}, y_{2} = \frac{2 λ D}{a}, y_{4} = \frac{4 λ D}{a}$
$y_{4} - y_{2} = \frac{2 λ D}{a} = 3 c m$
Width of central maxima is $\frac{2 λ D}{a} = 3 c m$

JEE - 2017

PHXII10:WAVE OPTICS

367728 In a double slit experiment, the two slits are 1 $m m$ apart and the screen is placed $1 m$ away. A monochromatic light wavelength $500 n m$ is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

1

0.1 m m

2

0.5 m m

3

0.02 m m

4

0.2 m m

Explanation:

Here, distance between two slits,
$d = 1 m m = 10^{- 3} m, D = 1 m$
$λ = 500 n m = 500 \times 10^{- 9} m$
width of central maxima in single slit diffraction pattern $\frac{2 λ D}{a}$
Fringe width in double slit experiment $β = \frac{λ D}{d}$
So, required condition $\frac{10 λ D}{d} = \frac{2 λ D}{a}$
$\Rightarrow a = \frac{d}{5} = \frac{1}{5} \times 10^{- 3} m = 0.2 m m$

PHXII10:WAVE OPTICS

367729 In a diffraction pattern due to a single slit of width $^{'} a^{'}$ , the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \overset{\circ}{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

1

\sin^{- 1} (\frac{1}{4})

2

\sin^{- 1} (\frac{2}{3})

3

\sin^{- 1} (\frac{1}{2})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

For first minima, $\sin 30^{\circ} = \frac{λ}{a} = \frac{1}{2}$
First secondary maxima will be at
$\sin θ = \frac{3 λ}{2 a} = \frac{3}{2} (\frac{1}{2}) \Rightarrow θ = \sin^{- 1} (\frac{3}{4})$

NEET - 2016

PHXII10:WAVE OPTICS

367726 Angular width of central maxima is $π / 2$ when a slit of width $^{'} A^{'}$ is illuminated by a light of wavelength $7000 \overset{\circ}{A}$ then $a =$

1

9 \times 10^{- 9} m

2

8.9 \times 10^{- 7} m

3

9 \times 10^{- 7} m

4

9.8 \times 10^{- 7} m

Explanation:

$\sin θ = \frac{λ}{a}, θ = 45^{\circ}, λ = 7000 \overset{o}{A}$
$a = \sqrt{2} λ = 9.8 \times 10^{- 7} m$

PHXII10:WAVE OPTICS

367727 A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $λ$ . If the second and fourth minima in the difdfraction pattern at distance $1 m$ from the slit are at $3 c m$ and $6 c m$ respectively from the central maximum what is the width of the central maximum? (i.e., distance between first minimum on either side of the central maximum)

1

4.5 c m

2

1.5 c m

3

6.0 c m

4

3.0 c m

Explanation:

The position of $n^{t h}$ minima is
$y = \frac{n λ D}{a}, y_{2} = \frac{2 λ D}{a}, y_{4} = \frac{4 λ D}{a}$
$y_{4} - y_{2} = \frac{2 λ D}{a} = 3 c m$
Width of central maxima is $\frac{2 λ D}{a} = 3 c m$

JEE - 2017

PHXII10:WAVE OPTICS

367728 In a double slit experiment, the two slits are 1 $m m$ apart and the screen is placed $1 m$ away. A monochromatic light wavelength $500 n m$ is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

1

0.1 m m

2

0.5 m m

3

0.02 m m

4

0.2 m m

Explanation:

Here, distance between two slits,
$d = 1 m m = 10^{- 3} m, D = 1 m$
$λ = 500 n m = 500 \times 10^{- 9} m$
width of central maxima in single slit diffraction pattern $\frac{2 λ D}{a}$
Fringe width in double slit experiment $β = \frac{λ D}{d}$
So, required condition $\frac{10 λ D}{d} = \frac{2 λ D}{a}$
$\Rightarrow a = \frac{d}{5} = \frac{1}{5} \times 10^{- 3} m = 0.2 m m$

PHXII10:WAVE OPTICS

367729 In a diffraction pattern due to a single slit of width $^{'} a^{'}$ , the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \overset{\circ}{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

1

\sin^{- 1} (\frac{1}{4})

2

\sin^{- 1} (\frac{2}{3})

3

\sin^{- 1} (\frac{1}{2})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

For first minima, $\sin 30^{\circ} = \frac{λ}{a} = \frac{1}{2}$
First secondary maxima will be at
$\sin θ = \frac{3 λ}{2 a} = \frac{3}{2} (\frac{1}{2}) \Rightarrow θ = \sin^{- 1} (\frac{3}{4})$

NEET - 2016

PHXII10:WAVE OPTICS

367726 Angular width of central maxima is $π / 2$ when a slit of width $^{'} A^{'}$ is illuminated by a light of wavelength $7000 \overset{\circ}{A}$ then $a =$

1

9 \times 10^{- 9} m

2

8.9 \times 10^{- 7} m

3

9 \times 10^{- 7} m

4

9.8 \times 10^{- 7} m

Explanation:

$\sin θ = \frac{λ}{a}, θ = 45^{\circ}, λ = 7000 \overset{o}{A}$
$a = \sqrt{2} λ = 9.8 \times 10^{- 7} m$

PHXII10:WAVE OPTICS

367727 A single slit of width $b$ is illuminated by a coherent monochromatic light of wavelength $λ$ . If the second and fourth minima in the difdfraction pattern at distance $1 m$ from the slit are at $3 c m$ and $6 c m$ respectively from the central maximum what is the width of the central maximum? (i.e., distance between first minimum on either side of the central maximum)

1

4.5 c m

2

1.5 c m

3

6.0 c m

4

3.0 c m

Explanation:

The position of $n^{t h}$ minima is
$y = \frac{n λ D}{a}, y_{2} = \frac{2 λ D}{a}, y_{4} = \frac{4 λ D}{a}$
$y_{4} - y_{2} = \frac{2 λ D}{a} = 3 c m$
Width of central maxima is $\frac{2 λ D}{a} = 3 c m$

JEE - 2017

PHXII10:WAVE OPTICS

367728 In a double slit experiment, the two slits are 1 $m m$ apart and the screen is placed $1 m$ away. A monochromatic light wavelength $500 n m$ is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?

1

0.1 m m

2

0.5 m m

3

0.02 m m

4

0.2 m m

Explanation:

Here, distance between two slits,
$d = 1 m m = 10^{- 3} m, D = 1 m$
$λ = 500 n m = 500 \times 10^{- 9} m$
width of central maxima in single slit diffraction pattern $\frac{2 λ D}{a}$
Fringe width in double slit experiment $β = \frac{λ D}{d}$
So, required condition $\frac{10 λ D}{d} = \frac{2 λ D}{a}$
$\Rightarrow a = \frac{d}{5} = \frac{1}{5} \times 10^{- 3} m = 0.2 m m$

PHXII10:WAVE OPTICS

367729 In a diffraction pattern due to a single slit of width $^{'} a^{'}$ , the first minimum is observed at an angle $30^{\circ}$ when light of wavelength $5000 \overset{\circ}{A}$ is incident on the slit. The first secondary maximum is observed at an angle of

1

\sin^{- 1} (\frac{1}{4})

2

\sin^{- 1} (\frac{2}{3})

3

\sin^{- 1} (\frac{1}{2})

4

\sin^{- 1} (\frac{3}{4})

Explanation:

For first minima, $\sin 30^{\circ} = \frac{λ}{a} = \frac{1}{2}$
First secondary maxima will be at
$\sin θ = \frac{3 λ}{2 a} = \frac{3}{2} (\frac{1}{2}) \Rightarrow θ = \sin^{- 1} (\frac{3}{4})$

NEET - 2016