A thin glass plate of thickness $$\frac{2500}{3}\lambda$$ ($$\lambda$$ is wavelength of light used) and refractive index $$\mu = 1.5$$ is inserted between one of the slits and the screen in Young's double slit experiment. At a point on the screen equidistant from the slits, the ratio of the intensities before and after the introduction of the glass plate is :

In Young's double slit experiment, without any glass plate, at a point equidistant from both slits, the path difference is zero. This results in constructive interference, and the intensity is maximum. Let the intensity from each slit be $$ I_0 $$. The resultant intensity is given by:

$$ I_{\text{before}} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos(0^\circ) = 2I_0 + 2I_0 \times 1 = 4I_0 $$

Now, a glass plate of thickness $$ t = \frac{2500}{3} \lambda $$ and refractive index $$ \mu = 1.5 $$ is inserted in front of one slit. This introduces an additional path difference for the light passing through that slit. The additional path difference $$ \Delta x $$ is calculated as:

$$ \Delta x = (\mu - 1) t = (1.5 - 1) \times \frac{2500}{3} \lambda = 0.5 \times \frac{2500}{3} \lambda = \frac{1}{2} \times \frac{2500}{3} \lambda = \frac{2500}{6} \lambda = \frac{1250}{3} \lambda $$

This path difference causes a phase difference $$ \phi $$, given by:

$$ \phi = \frac{2\pi}{\lambda} \times \Delta x = \frac{2\pi}{\lambda} \times \frac{1250}{3} \lambda = \frac{2\pi \times 1250}{3} = \frac{2500\pi}{3} $$

Since phase difference is periodic with $$ 2\pi $$, we reduce $$ \frac{2500\pi}{3} $$ modulo $$ 2\pi $$:

$$ \frac{2500\pi}{3} \div 2\pi = \frac{2500\pi}{3} \times \frac{1}{2\pi} = \frac{2500}{6} = \frac{1250}{3} $$

Now, $$ \frac{1250}{3} = 416 + \frac{2}{3} $$ (since $$ 416 \times 3 = 1248 $$, remainder $$ 1250 - 1248 = 2 $$). The fractional part is $$ \frac{2}{3} $$, so:

$$ \phi = \frac{2}{3} \times 2\pi = \frac{4\pi}{3} $$

After inserting the glass plate, the amplitudes from both slits remain the same (assuming no absorption), so the intensities are still $$ I_0 $$ for each slit. The resultant intensity is:

$$ I_{\text{after}} = I_0 + I_0 + 2\sqrt{I_0 I_0} \cos\left(\frac{4\pi}{3}\right) = 2I_0 + 2I_0 \cos\left(\frac{4\pi}{3}\right) $$

Since $$ \cos\left(\frac{4\pi}{3}\right) = \cos(240^\circ) = -\frac{1}{2} $$:

$$ I_{\text{after}} = 2I_0 + 2I_0 \times \left(-\frac{1}{2}\right) = 2I_0 - I_0 = I_0 $$

The ratio of intensities before and after inserting the glass plate is:

$$ \frac{I_{\text{before}}}{I_{\text{after}}} = \frac{4I_0}{I_0} = 4 : 1 $$

Hence, the correct answer is Option C.