JEE Main 2005 — Wave Optics Question with Solution

$$I=I_0{\left(\frac{\sin \theta }{\theta }\right)}^2$$

and $$\theta =\frac{\pi }{\lambda }\left(\frac{ay}{D}\right)$$

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 :

$\underset{\theta \to 0}{\lim }\left(\frac{\sin \theta }{\theta }\right)=1$

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.