Question

In a biprism experiment with sodium light, bands of width 0.0195 cm are observed at 100 cm from slit. On introducing a convex lens 30 cm away from the slit between biprism and screen, two images of the slit are seen 0.7 cm apart at 100 cm distance from the slit. Calculate the wavelength of sodium light.

• A. $\lambda =\text{4450 }\overset{^\circ }{A}$
• B. $\lambda =\text{5850 }\overset{^\circ }{A}$
• C. $\lambda =\text{2650 }\overset{^\circ }{A}$
• D. $\lambda =\text{6750 }\overset{^\circ }{A}$

Answer 1

Answer: (B)

Lens will from image of two sources at 70 cm from lens (i.e. screen)
From magnification of lens $m=\frac{v}{u}=\frac{h_{i}}{h_{o}}$
$\frac{70}{30}=\frac{0.7 / 2}{d / 2}$
On solving, $d=0.3cm$
Now, fringes width $\beta =\frac{\lambda D}{d}$
So, $\lambda =\frac{\beta d}{D}=\frac{0.0195 \times 10^{- 2} \times 0.3 \times 10^{- 2}}{1}=5850\overset{^\circ }{A}$