Physics - Wave Optics Question with Solution | TestHub

The position of the point exactly in front of one of the slits on the screen is $y=\frac{d}{2}$.

The path difference at this point is given by $\Delta x=\frac{yd}{D}=\frac{d^2}{2D}$.

For the third minimum, the path difference is $(2n-1)\frac{\lambda }{2}$ with $n=3$, which gives $\frac{5\lambda }{2}$.

Equating the path differences:
$\frac{d^2}{2D}=\frac{5\lambda }{2}\Rightarrow \frac{d^2}{D}=5\lambda$

Let the new wavelength be ${\lambda }^{\prime }$. For the first minimum at the same point, the path difference is $(2(1)-1)\frac{{\lambda }^{\prime }}{2}=\frac{{\lambda }^{\prime }}{2}$.

Equating the path differences again:
$\frac{d^2}{2D}=\frac{{\lambda }^{\prime }}{2}\Rightarrow {\lambda }^{\prime }=\frac{d^2}{D}$

Substituting the value of $\frac{d^2}{D}$ from the first case:
${\lambda }^{\prime }=5\lambda$

The required change in wavelength is:
$\Delta \lambda ={\lambda }^{\prime }-\lambda =5\lambda -\lambda =4\lambda$

Answer: $4\lambda$