Question 185: Waves Optics

Question

A beam of light consisting of wavelengths 650 nm and 550 nm illuminates the Young's double slits with separation of 2 mm such that the interference fringes are formed on a screen, placed at a distance of 1.2 m from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is ________ $	imes 10^{-5}$ m.

Accepted Answer

The wavelength of first light is $\lambda_1=650 \mathrm{~nm}=650 	imes 10^{-9} \mathrm{~m}$
The wavelength of second light is $\lambda_2=550 \mathrm{~nm}=550 	imes 10^{-9} \mathrm{~m}$
Slit separation is $\mathrm{d}=2 \mathrm{~mm}=2 	imes 10^{-3} \mathrm{~m}$
Screen distance is $\mathrm{D}=1.2 \mathrm{~m}$
For bright fringes of different wavelengths to coincide at the same distance $(y)$ from the central maximum, their positions must be equal :
$$ \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{~d}}=\frac{\mathrm{n}_2 \lambda_2 \mathrm{D}}{\mathrm{~d}} $$
$$ \mathrm{n}_1 \lambda_1=\mathrm{n}_2 \lambda_2 $$
Where $n_1$ and $n_2$ are the orders of the bright fringes for $\lambda_1$ and $\lambda_2$ respectively.
$$ \frac{\mathrm{n}_1}{\mathrm{n}_2}=\frac{\lambda_2}{\lambda_1}=\frac{550}{650}=\frac{11}{13} $$
Since we are looking for the least distance, we take the smallest integers that satisfy this ratio $\mathrm{n}_1=11$ and $\mathrm{n}_2=13$.
This means the $11^{	ext {th }}$ bright fringe of 650 nm coincides with the $13^{	ext {th }}$ bright fringe of 550 nm .
Using $\mathrm{n}_1=11$ and $\lambda_1=650 \mathrm{~nm}$ :
$$\Rightarrow $$ $$ \mathrm{y}=\frac{\mathrm{n}_1 \lambda_1 \mathrm{D}}{\mathrm{~d}} $$
$$\Rightarrow $$ $$ \mathrm{y}=\frac{11 	imes\left(650 	imes 10^{-9}ight) 	imes 1.2}{2 	imes 10^{-3}} \mathrm{~m}=429 	imes 10^{-5} \mathrm{~m} $$
The least distance from the central maximum where the bright fringes coincide is $429 	imes 10^{-5} \mathrm{~m}$.
Hence, the correct answer is 429 .

Waves Optics

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