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. $\lambda =4450\,\mathring{A}$
• 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, $d=0.3\,cm$
Now, fringes width $\beta =\frac{\lambda D}{d}$
So, $\lambda =\frac{\beta d}{D}=\frac{0.0195 \times 10^{- 2} \times 0.3 \times 10^{- 2}}{1}=5850\,\mathring{A}$