Question

In a Fresenel Biprism Experiment, the distance between the source and the screen is D and that between source and biprism is a. The wavelength of light used is l. The fringe width is b and refracting angle biprism is A. The refractive index of material of biprism is –

• A. 1 +$\frac{D\lambda}{2aA\beta}$
• B. 1 –$\frac{D\lambda}{2aA\beta}$
• C. 1 – $\frac{D\lambda}{aA\beta}$
• D. 1 +$\frac{aA}{D\lambda\beta}$

Answer 1

Answer: (A)

1 +$\frac{D\lambda}{2aA\beta}$

Answer 2

b = $\frac{D\lambda}{2a(\mu - 1)A}$ so m = 1 +$\frac{D\lambda}{2a(\mu - 1)A}$