Q. Angular width of central maxima in the Fraunhoffter diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength $6000\ \mathring{\mathrm{A}}$. When the slit is illuminated by light of another wavelength, the angular width decreases by 30%. The wavelength of this light will be
A) $6000\ \mathring{\mathrm{A}}$
B) $4200\ \mathring{\mathrm{A}}$
C) $3000\ \mathring{\mathrm{A}}$
D) $1800\ \mathring{\mathrm{A}}$
Explanation
(b) Angular width $\beta = \frac{2\lambda}{d} \Rightarrow \beta \propto \lambda$
$\Rightarrow \frac{\beta_{1}}{\beta_{2}} = \frac{\lambda_{1}}{\lambda_{2}} \Rightarrow \frac{\beta}{\frac{70}{100}\beta} = \frac{6000}{\lambda_{2}} \Rightarrow 4200\mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.