In a single slit diffraction the distance between plane of slit and screen is 1 ~m . The size of the slit is 0.7 ~mm and second maximum is formed at the distance of 2 ~mm from the centre of the screen, then find out the

Correct answer: A. 5600 Å For second maxima path difference, aligned & a θ=(5 λ)/(2) ; a · (y)/(D)=(5 λ)/(2) & λ=(2 a y)/(5 D)= 2 × 0.7 × 10^ -3 × 2 × 10^ -3 5 × 1 & =(2.8)/(5) × 10^ -6 ~m=(28)/(5) × 10^ -7 =5600 Å & aligned

AIIMS 201826 May

Exam: AIIMS
Year: 2018
Date: 26 May
Subject: Physics
Chapter: Wave Optics

AIIMS 2018 · 26 May · Physics · Wave Optics

In a single slit diffraction the distance between plane of slit and...

In a single slit diffraction the distance between plane of slit and screen is

1 \mathrm{~m}

. The size of the slit is

0.7 \mathrm{~mm}

and second maximum is formed at the distance of

2 \mathrm{~mm}

from the centre of the screen, then find out the wavelength of light.

Options

A $5600 Å$
B $2800 Å$
C $3000 Å$
D $6000 Å$

Correct answer

A. $5600 Å$

Step-by-step solution

For second maxima path difference,

\begin{aligned} & a \tan \theta=\frac{5 \lambda}{2} ; a \cdot \frac{y}{D}=\frac{5 \lambda}{2} \\ & \lambda=\frac{2 a y}{5 D}=\frac{2 \times 0.7 \times 10^{-3} \times 2 \times 10^{-3}}{5 \times 1} \\ & =\frac{2.8}{5} \times 10^{-6} \mathrm{~m}=\frac{28}{5} \times 10^{-7}=5600 Å \\ & \end{aligned}

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