Question

Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the 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)

For first diffraction min. $d \sin \theta=\lambda$
and if angle is small, $\sin \theta=\theta$
$\Rightarrow d \theta=\lambda$
Angular width of first dark fringe from central maxima is
$\theta=\frac{\lambda}{d}$
Full angular width of central maxima is
$w=2 \theta=\frac{2 \lambda}{d}$
Also we can use
$w^{\prime}=\frac{2 \lambda^{\prime}}{d}$
$\Rightarrow \frac{\lambda^{\prime}}{\lambda}=\frac{w^{\prime}}{w} $
$\Rightarrow \lambda^{\prime}=\lambda \frac{w^{\prime}}{w} $
$\Rightarrow \lambda^{\prime}=6000 \times 0.7=4200\,\mathring{A}$