Question

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

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

Answer 1

Answer: (B)

The condition for minima is given by
$ d = sin \, \theta = n \lambda$
For n = 1, we have
$ d \, sin \, \theta = \lambda$
If angle small, then $sin \, \theta = 0$
$\Rightarrow d \theta = \lambda$
Half angular width $\theta = \frac{\lambda}{d}$
Full angular width $2 \, \theta = 2 \frac{\lambda}{d}$
Also $\omega'$=$\frac{2\lambda}{d}^{'}$
$\therefore \frac{\lambda'}{\lambda} = \frac{\omega'}{\omega} \, \, or \, \, \lambda' = \lambda \frac{\omega'}{\omega}$
or $ \lambda' = 6000 \times 0.7$
$ = 4200 \mathring {A}$