Question

Light of wavelength $\lambda$ is incident on a slit. First minima of the diffraction pattern is found to lie at a distance of $6\, mm$ from the central maximum on a screen placed at a distance of $2 \,m$ from the slit. If slit width is $0.2\, mm$, then wavelength of the light used will be:

• A. $4000 \,\mathring{A}$
• B. $6000 \,\mathring{A}$
• C. $7000 \,\mathring{A}$
• D. $7400 \,\mathring{A}$

Answer 1

Answer: (B)

Location of diffraction minima in single slit diffraction pattern is given as
$ b \sin \theta=n \lambda $
$\Rightarrow b \cdot \frac{x}{D}=n \lambda$
$\Rightarrow \lambda=\frac{b x}{n D} $
$\Rightarrow \lambda=\frac{2 \times 10^{-4} \times 6 \times 10^{-3}}{1 \times 2}=6000 \,\mathring{A}$