Wave Optics

${\lambda _{obs}} = {\lambda _{source}}\sqrt {{{1 + {v \over C}} \over {1 - {v \over C}}}} $

For $v < < C$,

${{670.7} \over {670}} = 1 + {v \over C}$

$ \Rightarrow v = {{0.7} \over {670}} \times 3 \times {10^8}$ m/s

$ \Rightarrow v \simeq 3.13 \times {10^5}$ m/s

${{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} + {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}} } \over {{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} - {I_1} - {I_2} + 2\sqrt {{I_1}{I_2}} }}$

$ = {{2({I_1} + {I_2})} \over {4\sqrt {{I_1}{I_2}} }}$

$ = {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {2\sqrt {{{{I_1}} \over {{I_2}}}} }} = {{4 + 1} \over {2 \times 2}} = {5 \over 4}$

So $x = 4$

When electric field vector is completely removed and incident on Brewster's angle then only refraction takes place.

${I_P} = I + 9I + 2\sqrt {I \times 9I} \cos {\pi \over 2} = 10I$

${I_Q} = I + 9I + 2\sqrt {I \times 9I} \cos \pi = 4I$

So, ${I_P} - {I_Q} = 6I$

${{{I_{\max }}} \over {{I_{\min }}}} = {\left( {{{\sqrt {{I_1}} + \sqrt {{I_2}} } \over {\sqrt {{I_1}} - \sqrt {{I_2}} }}} \right)^2} = {\left( {{5 \over 1}} \right)^2}$

$ = {{25} \over 1}$

The answer is Option B : the distance between consecutive fringes will decrease.

Young's double-slit experiment depends on the principle of interference of light waves, and the resulting pattern of light and dark bands (fringes) depends on the wavelength of the light used.

The formula for the distance between fringes (y) in the double slit experiment is :

y = ${{\lambda D} \over d}$

where :

• λ is the wavelength of the light
• D is the distance from the slits to the screen
• d is the distance between the slits

The wavelength of orange light is approximately 600 nm while the wavelength of blue light is around 475 nm. Therefore, if you change the light from orange to blue, the wavelength λ decreases. As λ is directly proportional to y, if λ decreases, y will also decrease. So, the distance between consecutive fringes will decrease.

In the double-slit experiment, the position of a bright fringe is given by the formula :

$y = \frac{{m\lambda L}}{{d}}$,

where :

• m is the order of the fringe (1 for the first bright fringe, 2 for the second, etc.)
• $\lambda$ is the wavelength of the light (in meters)
• L is the distance from the slits to the screen (in meters)
• d is the distance between the slits (in meters)

We need to find the difference in position between the first and third bright fringes. So, we find the position of both and subtract the position of the first from the position of the third :

$y_3 = \frac{{3\lambda L}}{{d}}$

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

Then, the difference between the third and the first bright fringes is :

$y_3 - y_1 = 2\frac{{\lambda L}}{{d}}$

Now, let's plug the given values: $\lambda = 5890 \,Å = 5890 \times 10^{-10} \, m$ (since 1 Å = $10^{-10}$ meters), L = 0.5 m, and d = 0.5 mm = 0.5 $\times 10^{-3}$ m :

$ y_3 - y_1 = 2 \times \frac{5890 \times 10^{-10} \, \text{m} \times 0.5 \, \text{m}}{0.5 \times 10^{-3} \, \text{m}} = 1178 \times 10^{-6} \, \text{m} $

So, the correct answer is 1178 $\times 10^{-6}$ m, which corresponds to Option B.

Yes, you're absolutely correct. This equation you provided is for the angular size of the central maximum (or central diffraction disk) in a circular aperture diffraction pattern (like a pinhole) :

$ \sin \theta = \frac{1.22\lambda}{D} $

where :

• θ is the angle subtended by the radius of the central maximum at the pinhole,


• λ is the wavelength of the light, and


• D is the diameter of the aperture or pinhole.

If D (the diameter of the pinhole) is increased, then sinθ (and hence θ itself, for small θ) will decrease, implying that the size of the central maximum or diffraction disk will decrease. This is because less diffraction (bending of light) occurs when the pinhole is larger.

At the same time, increasing the size of the pinhole allows more light to pass through, which increases the intensity (brightness) of the light in the diffraction pattern.

So, increasing the diameter of the pinhole decreases the size of the diffraction pattern but increases its intensity, which confirms Option C.

As intensity of emergent beam is reduced to half after passing through two polarisers. It means angle between A and B is 0$^\circ$.

Now, on placing polariser C between A and B.

Intensity after passing through A is ${I_A} = {I \over 2}$.

Let $\theta$ be the angle between A and C. Intensity of light after passing through C is given by

${I_C} = {I \over 2}{\cos ^2}\theta $

Intensity of light after passing through polariser B is ${I \over 3}$.

Angle between C and B is also $\theta$ as A is parallel to B.

So, ${I \over 3} = {I_C}{\cos ^2}\theta = {I \over 2}{\cos ^2}\theta \,.\,{\cos ^2}\theta $

${\cos ^4}\theta = {2 \over 3} \Rightarrow \cos \theta = {\left( {{2 \over 3}} \right)^{1/4}}$

According to Malus law, we have

$I = {I_0}{\cos ^2}\theta \Rightarrow {I \over {{I_0}}} = {\cos ^2}\theta $

Since the observed intensities are same for all values of $\theta$, therefore, for the angles given in options, the observed intensity should be same as well for all $\theta$, $\theta$ = 8$^\circ$, 38$^\circ$, 188$^\circ$ and 218$^\circ$, ${I \over {{I_0}}} \sim 0.9$; and for $\theta$ = 203$^\circ$, ${I \over I} \sim 0.9$.

Therefore, angle between direction of polarisation and x axis is 203$^\circ$.

The slit width is a = 0.1 mm = 10^$-$4 m.

The wavelength of the light is $\lambda$ = 6000 $\times$ 10^$-$10 = 6 $\times$ 10^$-$7.

The distance from the slit to diffraction bands is D = 0.5 m.

We calculate the third dark band from the central band as follows:

$A\sin \theta = 3\lambda \Rightarrow \sin \theta = {{3\lambda } \over a} = {x \over D}$

$ \Rightarrow x = {{3\lambda D} \over a} = {{3 \times 6 \times {{10}^{ - 7}} \times 0.5} \over {{{10}^{ - 4}}}} = 9$ mm