Question

A single slit is illuminated with a parallel beam of light of wavelength 6000 $A^{\circ }$ and a diffraction pattern is obtained on a screen 1.2 m away from the slit. Distance between position of zero intensity on both side of central maxima is found to be 3.5 mm, then width of slit is

• A. 0.4 mm
• B. 0.8 mm
• C. 1.2 mm
• D. 0.7 mm

Answer 1

The correct option is A 0.4 mm

At minima,
$b\sin \theta =\lambda ,\text{and}\sin \theta =\tan \theta =\frac{D}{2x}$
$\Rightarrow b\times \frac{D}{2x}=\lambda \Rightarrow b=\frac{2\lambda x}{D}$
$b=\frac{2\times 6000\times {10}^{-10}\times 1.2}{3.5\times {10}^{-3}}=0.41mm$