The distances from a Fresnel biprism to a narrow slit and a screen are equal to $a = 25 ext{ cm}$

Question

The distances from a Fresnel biprism to a narrow slit and a screen are equal to $a = 25 	ext{ cm}$ and $b = 100 	ext{ cm}$ respectively. The refracting angle of the glass biprism is $	heta = 20'$ and its refractive index is $n = 1.5$. If the width of the fringe on the screen is $\Delta x = 0.55 	ext{ mm}$, find the wavelength of the light (in nm).

Professor PC Thomas Classes · Answer

The biprism produces two virtual coherent sources from the single slit. The deviation produced by each thin half of the biprism is $\delta = (n-1)	heta$. The distance between the two virtual sources is $d = 2a\delta = 2a(n-1)	heta$. The fringe width on the screen is given by $\Delta x = \frac{\lambda D}{d} = \frac{\lambda (a+b)}{2a(n-1)	heta}$. Solving for the wavelength $\lambda$, we get $\lambda = \frac{2a(n-1)	heta \Delta x}{a+b}$. First, convert the angle $	heta$ to radians: $	heta = 20' = \frac{20}{60} 	imes \frac{\pi}{180} = \frac{\pi}{540} 	ext{ rad}$. Substitute the values: $\lambda = \frac{2 	imes 0.25 	imes (1.5 - 1) 	imes (\pi/540) 	imes 0.55 	imes 10^{-3}}{0.25 + 1.00} = \frac{0.25 	imes (\pi/540) 	imes 0.55 	imes 10^{-3}}{1.25} = \frac{0.11 \pi}{540} 	imes 10^{-3} 	ext{ m} \approx 6.4 	imes 10^{-7} 	ext{ m} = 640 	ext{ nm}$.