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WAVE OPTICS

283436 In a Young's double slit experiment, distance between the two slits is $1 mm$ and the distance between the screen and the two slits is $1 m$ . If the fringe width on the screen is $0.06 cm$ then the wavelength of light is

1

1200 \AA

2

1000 \AA

3

6000 \AA

4

5000 \AA

Explanation:

: Given,
$d = 1 mm, D = 1 m$
$β = 0.06 cm = 0.06 \times 10^{- 2} m$
Fringe width, $β = \frac{λ D}{d}$
$λ = \frac{β d}{D}$
$= \frac{0.06 \times 10^{- 2} \times 1 \times 10^{- 3}}{1}$
$= 6 \times 10^{- 7} m$
$= 6000 \AA$

AMU-2004

WAVE OPTICS

283438 In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance $D$ of the screen from the slits would be made

1

2 D

2

D

3

\frac{D}{2}

4

\frac{D}{4}

Explanation:

: Given,
$d^{'} = 2 d$
$β^{'} = β$
Young's double-slit experiment the fringe width-
$β = \frac{λ D}{d}$
Where, $d =$ distance between slits
$D =$ distance between slits and screen
$λ =$ wavelength
$\frac{λ D^{'}}{d^{'}} = \frac{λ D}{d} (\begin{matrix} D^{'} = \\ \begin{matrix} New value of distance of \\ screen from double slit \end{matrix} \end{matrix})$
$\frac{D^{'}}{2 d} = \frac{D}{d}$
$D^{'} = 2 D$

UPSEE - 2010

WAVE OPTICS

283439 At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

1

\frac{π}{4}

radian

2

\frac{π}{2}

radian

3

π

radian

4

\frac{π}{8}

radian

Explanation:

:
original image
For the first minima at $P$
a $\sin θ = λ$
$\sin θ = \frac{Δ x}{a / 2}$
$Δ x = (a / 2) \sin θ$
Phase difference $(Δ ϕ) = \frac{2 π}{λ} \times Δ x$
$= \frac{2 π}{λ} \times \frac{a}{2} \sin θ (∵ a \sin θ = λ)$
$= \frac{2 π}{λ} \times \frac{λ}{2}$
$= π radian$

AIPMT - 2015

WAVE OPTICS

283440 In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle $30^{\circ}$ when of wavelength $5000 \AA$ is incident on the slit. The first secondary maxima is observed at an angle of

1

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

2

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

3

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

4

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

Explanation:

: For $n^{th}$ minima
$a \sin θ = n λ$
$a = slit width$
At $n = 1, θ = 30^{\circ}$
$a \sin 30^{\circ} = 1 \times λ$
$a = 2 λ$
For $n^{th}$ secondary maxima-
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For first secondary maxima-
$at n = 1$
a $\sin θ = \frac{3 λ}{2}$
$2 λ \times \sin θ = \frac{3 λ}{2}$
$\sin θ = \frac{3}{4}$
$θ = \sin^{- 1} (\frac{3}{4})$

NEET- 2016

WAVE OPTICS

283436 In a Young's double slit experiment, distance between the two slits is $1 mm$ and the distance between the screen and the two slits is $1 m$ . If the fringe width on the screen is $0.06 cm$ then the wavelength of light is

1

1200 \AA

2

1000 \AA

3

6000 \AA

4

5000 \AA

Explanation:

: Given,
$d = 1 mm, D = 1 m$
$β = 0.06 cm = 0.06 \times 10^{- 2} m$
Fringe width, $β = \frac{λ D}{d}$
$λ = \frac{β d}{D}$
$= \frac{0.06 \times 10^{- 2} \times 1 \times 10^{- 3}}{1}$
$= 6 \times 10^{- 7} m$
$= 6000 \AA$

AMU-2004

WAVE OPTICS

283438 In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance $D$ of the screen from the slits would be made

1

2 D

2

D

3

\frac{D}{2}

4

\frac{D}{4}

Explanation:

: Given,
$d^{'} = 2 d$
$β^{'} = β$
Young's double-slit experiment the fringe width-
$β = \frac{λ D}{d}$
Where, $d =$ distance between slits
$D =$ distance between slits and screen
$λ =$ wavelength
$\frac{λ D^{'}}{d^{'}} = \frac{λ D}{d} (\begin{matrix} D^{'} = \\ \begin{matrix} New value of distance of \\ screen from double slit \end{matrix} \end{matrix})$
$\frac{D^{'}}{2 d} = \frac{D}{d}$
$D^{'} = 2 D$

UPSEE - 2010

WAVE OPTICS

283439 At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

1

\frac{π}{4}

radian

2

\frac{π}{2}

radian

3

π

radian

4

\frac{π}{8}

radian

Explanation:

:
original image
For the first minima at $P$
a $\sin θ = λ$
$\sin θ = \frac{Δ x}{a / 2}$
$Δ x = (a / 2) \sin θ$
Phase difference $(Δ ϕ) = \frac{2 π}{λ} \times Δ x$
$= \frac{2 π}{λ} \times \frac{a}{2} \sin θ (∵ a \sin θ = λ)$
$= \frac{2 π}{λ} \times \frac{λ}{2}$
$= π radian$

AIPMT - 2015

WAVE OPTICS

283440 In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle $30^{\circ}$ when of wavelength $5000 \AA$ is incident on the slit. The first secondary maxima is observed at an angle of

1

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

2

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

3

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

4

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

Explanation:

: For $n^{th}$ minima
$a \sin θ = n λ$
$a = slit width$
At $n = 1, θ = 30^{\circ}$
$a \sin 30^{\circ} = 1 \times λ$
$a = 2 λ$
For $n^{th}$ secondary maxima-
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For first secondary maxima-
$at n = 1$
a $\sin θ = \frac{3 λ}{2}$
$2 λ \times \sin θ = \frac{3 λ}{2}$
$\sin θ = \frac{3}{4}$
$θ = \sin^{- 1} (\frac{3}{4})$

NEET- 2016

WAVE OPTICS

283436 In a Young's double slit experiment, distance between the two slits is $1 mm$ and the distance between the screen and the two slits is $1 m$ . If the fringe width on the screen is $0.06 cm$ then the wavelength of light is

1

1200 \AA

2

1000 \AA

3

6000 \AA

4

5000 \AA

Explanation:

: Given,
$d = 1 mm, D = 1 m$
$β = 0.06 cm = 0.06 \times 10^{- 2} m$
Fringe width, $β = \frac{λ D}{d}$
$λ = \frac{β d}{D}$
$= \frac{0.06 \times 10^{- 2} \times 1 \times 10^{- 3}}{1}$
$= 6 \times 10^{- 7} m$
$= 6000 \AA$

AMU-2004

WAVE OPTICS

283438 In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance $D$ of the screen from the slits would be made

1

2 D

2

D

3

\frac{D}{2}

4

\frac{D}{4}

Explanation:

: Given,
$d^{'} = 2 d$
$β^{'} = β$
Young's double-slit experiment the fringe width-
$β = \frac{λ D}{d}$
Where, $d =$ distance between slits
$D =$ distance between slits and screen
$λ =$ wavelength
$\frac{λ D^{'}}{d^{'}} = \frac{λ D}{d} (\begin{matrix} D^{'} = \\ \begin{matrix} New value of distance of \\ screen from double slit \end{matrix} \end{matrix})$
$\frac{D^{'}}{2 d} = \frac{D}{d}$
$D^{'} = 2 D$

UPSEE - 2010

WAVE OPTICS

283439 At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

1

\frac{π}{4}

radian

2

\frac{π}{2}

radian

3

π

radian

4

\frac{π}{8}

radian

Explanation:

:
original image
For the first minima at $P$
a $\sin θ = λ$
$\sin θ = \frac{Δ x}{a / 2}$
$Δ x = (a / 2) \sin θ$
Phase difference $(Δ ϕ) = \frac{2 π}{λ} \times Δ x$
$= \frac{2 π}{λ} \times \frac{a}{2} \sin θ (∵ a \sin θ = λ)$
$= \frac{2 π}{λ} \times \frac{λ}{2}$
$= π radian$

AIPMT - 2015

WAVE OPTICS

283440 In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle $30^{\circ}$ when of wavelength $5000 \AA$ is incident on the slit. The first secondary maxima is observed at an angle of

1

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

2

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

3

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

4

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

Explanation:

: For $n^{th}$ minima
$a \sin θ = n λ$
$a = slit width$
At $n = 1, θ = 30^{\circ}$
$a \sin 30^{\circ} = 1 \times λ$
$a = 2 λ$
For $n^{th}$ secondary maxima-
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For first secondary maxima-
$at n = 1$
a $\sin θ = \frac{3 λ}{2}$
$2 λ \times \sin θ = \frac{3 λ}{2}$
$\sin θ = \frac{3}{4}$
$θ = \sin^{- 1} (\frac{3}{4})$

NEET- 2016

WAVE OPTICS

283436 In a Young's double slit experiment, distance between the two slits is $1 mm$ and the distance between the screen and the two slits is $1 m$ . If the fringe width on the screen is $0.06 cm$ then the wavelength of light is

1

1200 \AA

2

1000 \AA

3

6000 \AA

4

5000 \AA

Explanation:

: Given,
$d = 1 mm, D = 1 m$
$β = 0.06 cm = 0.06 \times 10^{- 2} m$
Fringe width, $β = \frac{λ D}{d}$
$λ = \frac{β d}{D}$
$= \frac{0.06 \times 10^{- 2} \times 1 \times 10^{- 3}}{1}$
$= 6 \times 10^{- 7} m$
$= 6000 \AA$

AMU-2004

WAVE OPTICS

283438 In a Young's double slit experiment, the separation of the two slits is doubled. To keep the same spacing of fringes, the distance $D$ of the screen from the slits would be made

1

2 D

2

D

3

\frac{D}{2}

4

\frac{D}{4}

Explanation:

: Given,
$d^{'} = 2 d$
$β^{'} = β$
Young's double-slit experiment the fringe width-
$β = \frac{λ D}{d}$
Where, $d =$ distance between slits
$D =$ distance between slits and screen
$λ =$ wavelength
$\frac{λ D^{'}}{d^{'}} = \frac{λ D}{d} (\begin{matrix} D^{'} = \\ \begin{matrix} New value of distance of \\ screen from double slit \end{matrix} \end{matrix})$
$\frac{D^{'}}{2 d} = \frac{D}{d}$
$D^{'} = 2 D$

UPSEE - 2010

WAVE OPTICS

283439 At the first minimum adjacent to the central maximum of a single slit diffraction pattern, the phase difference between the Huygen's wavelet from the edge of the slit and the wavelet from the midpoint of the slit is

1

\frac{π}{4}

radian

2

\frac{π}{2}

radian

3

π

radian

4

\frac{π}{8}

radian

Explanation:

:
original image
For the first minima at $P$
a $\sin θ = λ$
$\sin θ = \frac{Δ x}{a / 2}$
$Δ x = (a / 2) \sin θ$
Phase difference $(Δ ϕ) = \frac{2 π}{λ} \times Δ x$
$= \frac{2 π}{λ} \times \frac{a}{2} \sin θ (∵ a \sin θ = λ)$
$= \frac{2 π}{λ} \times \frac{λ}{2}$
$= π radian$

AIPMT - 2015

WAVE OPTICS

283440 In a diffraction pattern due to a single slit of width a, the first minima is observed at an angle $30^{\circ}$ when of wavelength $5000 \AA$ is incident on the slit. The first secondary maxima is observed at an angle of

1

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

2

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

3

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

4

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

Explanation:

: For $n^{th}$ minima
$a \sin θ = n λ$
$a = slit width$
At $n = 1, θ = 30^{\circ}$
$a \sin 30^{\circ} = 1 \times λ$
$a = 2 λ$
For $n^{th}$ secondary maxima-
$a \sin θ = (\frac{2 n + 1}{2}) λ$
For first secondary maxima-
$at n = 1$
a $\sin θ = \frac{3 λ}{2}$
$2 λ \times \sin θ = \frac{3 λ}{2}$
$\sin θ = \frac{3}{4}$
$θ = \sin^{- 1} (\frac{3}{4})$

NEET- 2016

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