Question

Monochromatic light with wavelength 690 nm is incident on a slit with width 0.0329 mm. The distance from the slit to a screen is 3.2 m. Consider a point on the screen 1.3 cm from the central maximum. Calculate (a) θ for that point, (b) α, and (c) the ratio of the intensity at that point to the intensity at the central maximum.

Answer 1

Step-by-step explanation:

We have given,

$Wavelength\ of the\ light(\lambda)} = 690mm = 690 \times 10^{-9}m$

$\text {width of the slit(w)} = 0.0329\ mm = 0.0329 \times 10^{-3}m$

distance of the slit from the screen(d) = 3.2m

distance from the central maximum(m) = 1.3cm = 0.013m

Now,

(a). The expression for the angle of the inclination of the wave is ($\theta$ )

$\theta = sin^{-1} \dfrac md\\\theta = sin^{-1}( \dfrac {0.013}{3.2})\\\\\theta = sin^{-1}(0.0040625)\\\\\theta = sin^{-1}(sin 0.23^o)$

Thus the angle of inclination of the wave is

$\theta= 0.23^o$

(b). The expression for the angle $\alpha$

$\alpha = \dfrac {\pi \times w}{\lambda} sin \theta\\\alpha = \dfrac {3.14 \times 0.0329 \times 10^{-3}}{690 \times 10^{-9}} sin {0.23^o}\\\\\alpha = \dfrac {3.14 \times 0.0329 \times 10^{-3} \times 0.004.625}{690 \times 10^{-9}}$

after solving the equation we get,

Thus the angle $\alpha$ = 0.61 radian

(c). the expression for the ratio of the intensity is

$\dfrac{I_0}{I} = (\dfrac {sin \alpha}{\alpha}) ^2$

$\dfrac{I_0}{I} = (\dfrac {sin (0.61)}{0.61})^2\\\dfrac{I_0}{I} = (\dfrac {0.0.65}{0.61})^2\\\dfrac{I_0}{I} = 0.0305$

Thus the ratio of the intensity is 0.0305