Question 390: Wave Optics

Question

In biprism experiment, the distance between source and eyepiece is 1.2 m, the distance between two virtual sources is 0.84 mm. Then the wavelength of light used if eyepiece is to be moved transversely through a distance of 2.799 cm to shift 30 fringes is

Accepted Answer

In a biprism experiment, wavelength of the light is given as
$$\lambda=\frac{d \beta}{D}\quad 	ext{.... (i)}$$
where, $\beta$ is the fringe width,
$d$ is the distance between the two sources and $D$ is the distance between the source and eyepeice.
Given, $D=1.2 \mathrm{~m}$,
$$\begin{aligned}
& d=0.84 \mathrm{~mm}=0.84 	imes 10^{-3} \mathrm{~m} 	ext { and } \
& \beta=\frac{2.799 	imes 10^{-2}}{30}=9.33 	imes 10^{-4}
\end{aligned}$$
Substituting these values in Eq. (i), we get
$$\begin{aligned}
\lambda & =\frac{0.84 	imes 10^{-3} 	imes 9.33 	imes 10^{-4}}{1.2} \
& =6.531 	imes 10^{-7} \mathrm{~m}=6531 \mathop A\limits^o
\end{aligned}$$

Answer

6533 $$\mathop A\limits^o $$

Answer

6537 $$\mathop A\limits^o $$

Answer

6535 $$\mathop A\limits^o $$

Answer

6351 $$\mathop A\limits^o $$

Wave Optics

In biprism experiment, the distance between source and eyepiece is 1.2 m, the distance between two virtual sources is 0.84 mm. Then the wavelength of light used if eyepiece is to be moved transversely through a distance of 2.799 cm to shift 30 fringes is

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