Wave Optics

$ \begin{aligned} &\text { According to Brewster's law }\\ &\begin{aligned} & & \mu_{g o} & =\frac{\mu_g}{\mu_w}=\tan i_p=\sqrt{3} \\ \therefore & & \mu_g & =\sqrt{3} \cdot \sqrt{2}=\sqrt{6} \end{aligned} \end{aligned} $

For 5th bright fringe

$ x^5=n \frac{D \lambda}{d}=\frac{5 \times 1 \times \lambda}{1 \times 10^{-3}} $

and 3rd dark fringe

$ x^3=\frac{(2 n-1) D \lambda}{2 d}=\frac{5 \times 1 \times \lambda}{2 \times 1 \times 10^{-3}} $

Now, $x^5-x^3=1.63 \mathrm{~mm}$

$ \begin{aligned} \frac{5 \lambda}{10^{-3}}-\frac{5 \lambda}{2 \times 10^{-3}} & =1.63 \times 10^{-3} \\ \frac{5 \lambda}{2} & =1.63 \times 10^{-3} \\ \lambda & =\frac{3.26}{5} \times 10^{-6}=6.5 \times 10^{-7} \mathrm{~m} \end{aligned} $

In Young's double slit experiment within a medium with a refractive index of 1.33, the maximum intensity is represented as $ I_0 $.

To find the intensity at a point on the screen where the path difference between the light from the slits is $ \frac{\lambda}{4} $, follow these steps:

Path Difference: Given as $ \Delta x = \frac{\lambda}{4} $.

Phase Difference Calculation:

$ \phi = \left(\frac{2\pi}{\lambda}\right) \times \Delta x $

Substituting the given path difference:

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

Intensity Calculation: The intensity at any point on the screen with a phase difference $ \phi $ is given by:

$ I = I_0 \cos^2\left(\frac{\phi}{2}\right) $

Substituting the phase difference:

$ I = I_0 \cos^2\left(\frac{\pi/2}{2}\right) $

Simplifying further:

$ I = I_0 \cos^2\left(\frac{\pi}{4}\right) $

$ I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 $

$ I = \frac{I_0}{2} $

Thus, the intensity at the specified point on the screen is $\frac{I_0}{2}$.

In the Young's Double Slit Experiment (YDSE), for monochromatic light shining on a screen 1.80 m away from two slits separated by 2.08 mm, the separation of the first and second order bright fringes is 0.553 mm. To find the wavelength of light used, we use the following calculations:

The position of the first bright fringe is given by:

$ y_1 = \frac{\lambda D}{d} $

The position of the second bright fringe is:

$ y_2 = \frac{2\lambda D}{d} $

The separation between the first and second order bright fringes is:

$ y_2 - y_1 = \frac{2\lambda D}{d} - \frac{\lambda D}{d} = \frac{\lambda D}{d} $

From this, we can derive the formula for the wavelength $\lambda$ as:

$ \lambda = \frac{(y_2 - y_1) d}{D} $

Substituting the given values into this equation:

$ \lambda = \frac{0.553 \times 10^{-3} \times 2.08 \times 10^{-3}}{1.8} $

Solving this gives:

$ \lambda = 639 \times 10^{-9} \text{ m} $

Therefore, the wavelength of the light used is:

$ \lambda = 639 \text{ nm} $

On introduction of a transparent sheet,

$\Delta x=(\mu-1) t$

if the central maxima occupies position of the nth fringe, then $(\mu-1) t=n \lambda=d \sin \theta$

$\begin{aligned} \Rightarrow \quad \sin \theta & =\frac{(\mu-1)}{d} t \\ & =\frac{(1.25-1) \times 1.5 \times 10^{-7}}{3 \times 10^{-7}}=0.125 \end{aligned}$

Therefore, central maxima has angular position, $\theta=\sin ^{-1}(0.125)=7.18^{\circ} \approx 7.2^{\circ}$

for small angles, $\sin \theta \approx \tan \theta$

$\begin{aligned} \tan \theta & =\frac{y}{D} \\ \therefore \quad \frac{y}{D} & =\frac{(\mu-1) t}{d} \Rightarrow y=\frac{D(\mu-1)}{d} t \end{aligned}$

As, shift $\Delta x = {\beta \over \lambda }(\mu - 1)t$

Shift due to one plate $\Delta {x_1} = {\beta \over \lambda }({\mu _1} - 1)t$

Shift due to another path $\Delta {x_2} = {\beta \over \lambda }({\mu _2} - 1)t$

Net shift $\Delta x = \Delta {x_2} - \Delta {x_1} = {\beta \over \lambda }({\mu _2} - {\mu _1})t$ ..... (i)

Also, it is given that $\Delta$x = 6$\beta$

Hence, $6\beta = {\beta \over \lambda }({\mu _1} - {\mu _2})t$

$ \Rightarrow t = {{6\lambda } \over {({\mu _1} - {\mu _2})}} = {{6 \times 4000} \over {(1.7 - 1.4)}} - 80000\,\mathop A\limits^o $

$ \Rightarrow t = 8 \times {10^{ - 6}} = 8\mu $m