Question

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

• A. $\lambda =4450\mathring{A}$
• B. $\lambda =5850\mathring{A}$
• C. $\lambda =2650\mathring{A}$
• D. $\lambda =6750\mathring{A}$

Answer 1

Answer: (B)

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