JEE MAIN - Physics (2005 - No. 21)
If $${I_0}$$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?
$$4{I_0}$$
$$2{I_0}$$
$${{{I_0}} \over 2}$$
$${I_0}$$
Explanation
$$I = {I_0}{\left( {{{\sin \theta } \over \theta }} \right)^2}$$
and $$\theta = {\pi \over \lambda }\left( {{{ay} \over D}} \right)$$
For principal maximum y = 0
$$ \therefore $$ $$\theta $$ = 0
When $\theta = 0$, the intensity formula becomes :
$I = I_0\left(\frac{\sin \theta}{\theta}\right)^2 = I_0\left(\frac{\sin 0}{0}\right)^2$
Here, we have an indeterminate form $\frac{0}{0}$. However, by taking the limit of the expression as $\theta$ approaches 0, we find that :
$\lim\limits_{\theta \to 0} \left(\frac{\sin \theta}{\theta}\right) = 1$
Therefore, the intensity at the principal maximum remains the same :
$I = I_0\left(\frac{\sin \theta}{\theta}\right)^2 = I_0(1)^2 = I_0$
Hence intensity will remain same.
and $$\theta = {\pi \over \lambda }\left( {{{ay} \over D}} \right)$$
For principal maximum y = 0
$$ \therefore $$ $$\theta $$ = 0
When $\theta = 0$, the intensity formula becomes :
$I = I_0\left(\frac{\sin \theta}{\theta}\right)^2 = I_0\left(\frac{\sin 0}{0}\right)^2$
Here, we have an indeterminate form $\frac{0}{0}$. However, by taking the limit of the expression as $\theta$ approaches 0, we find that :
$\lim\limits_{\theta \to 0} \left(\frac{\sin \theta}{\theta}\right) = 1$
Therefore, the intensity at the principal maximum remains the same :
$I = I_0\left(\frac{\sin \theta}{\theta}\right)^2 = I_0(1)^2 = I_0$
Hence intensity will remain same.
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