Question

The angular width of central maxima in the Fraunhofer diffraction pattern is measured. The slit is illuminated by the light of wavelength $6000\mathring{A}$ If the slit is illuminated by a light of another wavelength, angular width decreases by 30%. The wavelength of light used is :

• A. $3500\mathring{A}$
• B. $4200\mathring{A}$
• C. $4700\mathring{A}$
• D. $6000\mathring{A}$

Answer 1

The correct option is B $4200\mathring{A}$

$d\sin \theta =n\lambda$
n = 1, we have
$d\sin \theta =\lambda$
If angle is small, then $\sin \theta =\theta$
$d\theta =\lambda$
Half angular width $\theta =\frac{\lambda }{d}$
Full angular width $2\theta =2\frac{\lambda }{d}$
${\omega }^{\prime }=\frac{2\lambda }{d}$
$\frac{{\lambda }^{\prime }}{\lambda }=\frac{{\omega }^{\prime }}{\omega }\Rightarrow {\lambda }^{\prime }={\lambda }^{\prime }\frac{{\omega }^{\prime }}{omega}$
${\lambda }^{\prime }=6000\times 0.7=4200\mathring{A}$