91影视

wavelength A= 620 nm, distance from the screen D = 3 m, and width of the slit are d=0.450 mm. The intensity at the central maximum is$l_0$

Let x be the distance of a point with the intensity I from the central maximum. Then we can represent this intensity as.

$I=I_0{\left[\frac{(\sin {\displaystyle \frac{\left(蟺d\right)\sin \left(胃\right)}{位}}}{\frac{\left(蟺d\right)\sin \left(胃\right)}{位}}\right]}^2$

Now the relationship between x and D can be expressed in terms of$\sin 胃$ as

$\sin 胃=\frac{x}{D}$

For a very small angle, we can write$胃$ as $胃$. So the above equation becomes

$胃=\frac{x}{D}$

When the distance from the central maximum is x =1 mm, by substituting the value of in terms of intensity, we get

$$\begin{gathered} I=I_0{\left[\frac{(\sin {\displaystyle \frac{\left(蟺d\right)\sin \left(胃\right)}{位}}}{\frac{\left(蟺d\right)\sin \left(胃\right)}{位}}\right]}^2 \\ I=I_0{\left[\frac{(\sin {\displaystyle \frac{(蟺0.450脳{10}^{-3})(1脳{10}^{-3})}{\left(620脳{10}^{-9}\right)\left(3\right)}}}{\frac{(蟺0.450脳{10}^{-3})(1脳{10}^{-3})}{\left(620脳{10}^{-9}\right)\left(3\right)}}\right]}^2 \\ =I_0{\left[\frac{\sin \left(0.760\right)}{0.760}\right]}^2 \\ =0.8221l_0 \end{gathered}$$

When the distance from the central maximum is x = 3 mm, by substituting this value in the above intensity equation, we get

$$\begin{gathered} I=I_0{\left[\frac{(\sin {\displaystyle \frac{(蟺0.450脳{10}^{-3})(1脳{10}^{-3})}{\left(620脳{10}^{-9}\right)\left(3\right)}}}{\frac{(蟺0.450脳{10}^{-3})(1脳{10}^{-3})}{\left(620脳{10}^{-9}\right)\left(3\right)}}\right]}^2 \\ =I_0{\left[\frac{\sin \left(2.27\right)}{2.27}\right]}^2 \\ =0.1131l_0 \end{gathered}$$

When the distance from the central maximum is x=5 mm, by substituting this value in the above intensity equation, we get

$$\begin{gathered} I=I_0{\left[\frac{(\sin {\displaystyle \frac{(蟺0.450脳{10}^{-3})(1脳{10}^{-3})}{\left(620脳{10}^{-9}\right)\left(3\right)}}}{\frac{(蟺0.450脳{10}^{-3})(1脳{10}^{-3})}{\left(620脳{10}^{-9}\right)\left(3\right)}}\right]}^2 \\ =I_0{\left[\frac{\sin \left(3.79\right)}{3.79}\right]}^2 \\ =0.0253l_0 \end{gathered}$$