Question

In a biprism experiment by using light of wavelength $5000\overset{\text{o}}{\mathop{\text{A}}}\,,$ 5 mm wide fringes are obtained on a screen 1.0 m away from the coherent sources. The separation between the two coherent sources is

• A. 1.0 mm
• B. 0.1 mm
• C. 0.05 mm
• D. 0.01 mm

Answer 1

Answer: (B)

0.1 mm

Answer 2

Given, $\lambda =5000\,\,\overset{\text{o}}{\mathop{\text{A}}}\,=5\times {{10}^{-7}}m.$ $\beta =5mm=5\times {{10}^{-3}}m$ . $D=1\,\,m$ . $d=?$ Fringe width $\beta =\frac{D\lambda }{d}$ $\therefore$ $d=\frac{D\lambda }{\beta }=\frac{1\times 5\times {{10}^{-7}}}{5\times {{10}^{-3}}}$ $=1\times {{10}^{-4}}m=0.1\,\,mm$