Question 14: Wave Optics

Question

In a Young's double slit experiment for a particular wavelength of light the distance between third dark and fifth bright fringe is 1.63 mm . If the distance between two slit is 1 mm and distance between slit and screen is 1 metre. Then, the wavelength of light is

Accepted Answer

For 5th bright fringe
$$ x^5=n \frac{D \lambda}{d}=\frac{5 	imes 1 	imes \lambda}{1 	imes 10^{-3}} $$
and 3rd dark fringe
$$ x^3=\frac{(2 n-1) D \lambda}{2 d}=\frac{5 	imes 1 	imes \lambda}{2 	imes 1 	imes 10^{-3}} $$
Now, $x^5-x^3=1.63 \mathrm{~mm}$
$$ \begin{aligned} \frac{5 \lambda}{10^{-3}}-\frac{5 \lambda}{2 	imes 10^{-3}} & =1.63 	imes 10^{-3} \ \frac{5 \lambda}{2} & =1.63 	imes 10^{-3} \ \lambda & =\frac{3.26}{5} 	imes 10^{-6}=6.5 	imes 10^{-7} \mathrm{~m} \end{aligned} $$

Answer

$5.5 	imes 10^{-7} \mathrm{~m}$

Answer

$6 	imes 10^{-7} \mathrm{~m}$

Answer

$6.5 	imes 10^{-7} \mathrm{~m}$

Answer

$7.5 	imes 10^{-7} \mathrm{~m}$

Wave Optics

In a Young's double slit experiment for a particular wavelength of light the distance between third dark and fifth bright fringe is 1.63 mm . If the distance between two slit is 1 mm and distance between slit and screen is 1 metre. Then, the wavelength of light is

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