Question

A screen is placed $50\, cm$ from a single slit which is illuminated with light of wavelength $6000 ?$. If the distance between the first and third minima in the diffraction pattern is $3.0 \,mm$. The width of the slit is

• A. $1 \times 10^{-4} \,m$
• B. $2 \times 10^{-4} \,m$
• C. $0.5 \times 10^{-4} \,m$
• D. $4 \times 10^{-4} \,m$

Answer 1

The Correct Option is B

The position of $n^{th}$ minima in the diffraction pattern is $x_{n} = \frac{nD\lambda}{d} $ $ \therefore x_{3} -x_{1} = \left(3-1\right) \frac{D\lambda}{d} $ $= \frac{2D\lambda}{d} $ or $d =\frac{ 2D\lambda}{x_{3} -x_{1} }$ $ =\frac{ 2\times 0.50 \times 6000 \times 10^{-10}}{3\times 10^{-3}}$ $ = 2 \times 10^{-4} m $