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
$dsin\theta =\lambda$
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 ${\omega }^{\prime }$=${\frac{2\lambda }{d}}^{{}^{\prime }}$
$\therefore \frac{{\lambda }^{\prime }}{\lambda }=\frac{{\omega }^{\prime }}{\omega }or{\lambda }^{\prime }=\lambda \frac{{\omega }^{\prime }}{\omega }$
or ${\lambda }^{\prime }=6000\times 0.7$
$=4200\mathring{A}$