Answer to Question #146734 in Optics for Promise Omiponle

Question

A laser beam is incident on two slits with separation d = 0.026 mm. A screen is placed L = 3.2 m from the slits. The wavelength of the laser light is λ = 4750 Å. θ1 and θ2 are the angles to the first and second bright fringes above the center of the screen.

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(a) Express sin(θ1) in terms of d and λ.
(b) Express sin(θ2) in terms of d and λ.
(c) Express the distance between the two bright fringes on the screen, y, in terms of θ1, θ2 and L.
(d) Solve for the numerical value of y in meters.

Answer 1

As per the given question,

Separation between the slits $(d)=0.026mm=2.6\times 10^{-5}m$

Screen distance $(L)=3.2m$

Wavelength of the light $(\lambda)=4750A^\circ = 0.0475\times 10^{-5}m$

For the central maxima $y=\frac{n\lambda L}{2d}$

for the first maxima, n=1

$y_1=\frac{\lambda L}{2d}$

for the second n = 2

$y_2=\frac{2\lambda L}{2d}=\frac{\lambda L}{d}$

Hence $\sin \theta_1=\frac{y_1}{\sqrt{L^2+y_1^2}}$

$\Rightarrow \sin \theta_1=\frac{\frac{\lambda L}{2d}}{\sqrt{L^2+(\frac{\lambda L}{2d})^2}}$

$\Rightarrow \sin \theta_1=\frac{\lambda }{\sqrt{4d^2+\lambda ^2}}$

$b)$ $\sin \theta_2=\frac{y_2}{\sqrt{L^2+y_2^2}}$

$\Rightarrow \sin \theta_2=\frac{\frac{\lambda L}{d}}{\sqrt{L^2+(\frac{\lambda L}{d})^2}}$

$\Rightarrow \sin \theta_2=\frac{\lambda }{\sqrt{d^2+\lambda ^2}}$

$c)$ The distance between the two bright fringes on the screen,

$y=y_2-y_1$

$=L\tan\theta_2 -L\tan\theta_1$

$=L(\tan\theta_2-\tan\theta_1)$

$d)$ $y=y_2-y_1$

$=\frac{\lambda L}{d}-\frac{\lambda L}{2d}$

$=\frac{\lambda L}{2d}$

$=\frac{0.0475\times 10^{-5}m\times 3.2m}{2\times 2.6\times 10^{-5}m}$

$=0.029m$