Refraction of Light Questions in English

(A) Given,angle of incidence $i = 2r$,where $r$ is the angle of refraction.
According to Snell's law,$\mu_1 \sin i = \mu_2 \sin r$.
Assuming the light travels from air $(\mu_1 = 1)$ to a medium with refractive index $\mu_2 = \mu$,we have:
$1 \cdot \sin(2r) = \mu \sin r$
Using the trigonometric identity $\sin(2r) = 2 \sin r \cos r$,we get:
$2 \sin r \cos r = \mu \sin r$
Since $\sin r \neq 0$,we can divide both sides by $\sin r$:
$2 \cos r = \mu \implies \cos r = \frac{\mu}{2}$
Thus,$r = \cos^{-1}\left(\frac{\mu}{2}\right)$.
Since $i = 2r$,the angle of incidence is $i = 2 \cos^{-1}\left(\frac{\mu}{2}\right)$.

(A) According to Snell's Law,the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media:
$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{\mu_2}{\mu_1}$
We also know that the refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ in the medium,given by $\mu = \frac{c}{v} = \frac{\lambda_0}{\lambda}$,where $\lambda_0$ is the wavelength in vacuum.
Therefore,$\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$.
Equating the two expressions,we get:
$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{\lambda_1}{\lambda_2}$
Rearranging for $\lambda_2$,we find:
$\lambda_2 = \lambda_1 \frac{\sin \alpha_2}{\sin \alpha_1}$

(D) The air bubble is located at the centre $O$ of the solid glass sphere.
When light rays originating from the air bubble travel towards the surface of the sphere,they move along the radius of the sphere.
Since the radius is always perpendicular to the surface of the sphere,the light rays strike the surface at an angle of incidence $i = 0^\circ$.
According to Snell's law,when light is incident normally on a surface,it does not undergo any refraction or deviation.
Therefore,the light rays continue to travel in a straight line without bending.
As a result,the observer looking from outside sees the image of the bubble at the same position as the object.
Thus,the apparent distance of the bubble from the centre of the sphere is zero.

(B) From Snell's law,we have $\mu_X \sin i = \mu_Y \sin r$.
Since $\mu = \frac{c}{V}$,we can write $\frac{c}{V_X} \sin i = \frac{c}{V_Y} \sin r$,which simplifies to $\frac{\sin i}{\sin r} = \frac{V_X}{V_Y}$.
From the given graph,the slope of the line is $\tan 30^{\circ} = \frac{\sin r}{\sin i} = \frac{1}{\sqrt{3}}$.
Therefore,$\frac{\sin i}{\sin r} = \sqrt{3}$.
Substituting this into the velocity ratio,we get $\frac{V_X}{V_Y} = \sqrt{3}$,which implies $V_X = \sqrt{3} V_Y$.
Since the light travels from a rarer medium $(X)$ to a denser medium $(Y)$ (as $r < i$),total internal reflection cannot occur when light is incident from medium $X$ to $Y$. Frequency remains constant during refraction.

(B) According to the law of reflection,the angle of incidence $i$ is equal to the angle of reflection $\theta$. So,$i = \theta$.
The sum of the angles on a straight line is $180^{\circ}$. Therefore,the angle of reflection $\theta$,the angle between the reflected and refracted rays $(90^{\circ})$,and the angle of refraction $r$ must satisfy:
$\theta + 90^{\circ} + r = 180^{\circ}$
Substituting $\theta = i$:
$i + 90^{\circ} + r = 180^{\circ}$
$r = 90^{\circ} - i$
According to Snell's law,$\mu = \frac{\sin i}{\sin r}$.
Substituting $r = 90^{\circ} - i$:
$\mu = \frac{\sin i}{\sin(90^{\circ} - i)}$
Since $\sin(90^{\circ} - i) = \cos i$,we get:
$\mu = \frac{\sin i}{\cos i} = \tan i$
Thus,$\tan i = \mu$.

(C) When light travels from one medium to another,its speed and wavelength change because the optical density of the medium changes. However,the frequency of light is determined by the source of the light and remains constant during refraction. Therefore,the frequency does not change.

(D) According to Snell's law,for a series of parallel layers,the product of the refractive index and the sine of the angle of incidence remains constant at every interface.
Let $\mu_0$ be the refractive index of the $0^{\text{th}}$ layer and $i_0$ be the angle of incidence at the first interface.
$\mu_0 \sin i_0 = \mu_m \sin r_m$
Given:
$\mu_0 = \mu = 1.5$
$i_0 = 30^{\circ}$
$\mu_m = \mu - m \Delta \mu = 1.5 - m(0.015)$
Since the ray emerges parallel to the interface after the $m^{\text{th}}$ refraction,the angle of refraction $r_m = 90^{\circ}$.
Substituting the values:
$1.5 \sin 30^{\circ} = (1.5 - m \times 0.015) \sin 90^{\circ}$
$1.5 \times 0.5 = (1.5 - 0.015m) \times 1$
$0.75 = 1.5 - 0.015m$
$0.015m = 1.5 - 0.75$
$0.015m = 0.75$
$m = \frac{0.75}{0.015} = \frac{750}{15} = 50$
Thus,the value of $m$ is $50$.

(D) The wavelength of light in a medium is given by the formula $\lambda_m = \frac{\lambda_0}{n}$,where $\lambda_0$ is the wavelength in vacuum (or air) and $n$ is the refractive index of the medium.
Given:
$\lambda_0 = 4200 Å$
$n = 4/3$
Substituting the values:
$\lambda_m = \frac{4200}{4/3} = 4200 \times \frac{3}{4} = 1050 \times 3 = 3150 Å$.
Therefore,the wavelength of the light in water is $3150 Å$.

(C) The frequency $f$ of light remains constant when it travels from one medium to another.
Since $v = f \lambda$ and $v = \frac{c}{\mu}$,we have $\frac{c}{\mu} = f \lambda$,which implies $\mu \lambda = \frac{c}{f} = \text{constant}$.
Therefore,$\mu_1 \lambda_1 = \mu_2 \lambda_2$.
Given: $\mu_1 = \frac{4}{3}$,$\lambda_1 = 540 \ nm$,and $\mu_2 = \frac{3}{2}$.
Substituting the values: $\left(\frac{4}{3}\right) \times 540 = \left(\frac{3}{2}\right) \times \lambda_2$.
$\lambda_2 = 540 \times \frac{4}{3} \times \frac{2}{3}$.
$\lambda_2 = 540 \times \frac{8}{9} = 60 \times 8 = 480 \ nm$.

(C) The speed of light in a vacuum is $c = 3 \times 10^8 \text{ m/s}$.
The speed of light in the medium is $v = 200 \times 10^8 \text{ cm/s}$.
First,convert the speed in the medium to $\text{m/s}$:
$v = 200 \times 10^8 \times 10^{-2} \text{ m/s} = 2 \times 10^8 \text{ m/s}$.
The refractive index $n$ is given by the formula $n = \frac{c}{v}$.
Substituting the values: $n = \frac{3 \times 10^8 \text{ m/s}}{2 \times 10^8 \text{ m/s}} = 1.5$.

(A) From Snell's law,$\mu_1 \sin \alpha = \mu_2 \sin \beta$.
Given $\mu_1 = 1$ and $\mu_2 = 1.5$.
The vector $\vec{AO} = 2\hat{i} - 3\hat{j}$ makes an angle $\alpha$ with the vertical axis (y-axis). Thus,$\tan \alpha = \frac{|x|}{|y|} = \frac{2}{3}$.
Therefore,$\sin \alpha = \frac{2}{\sqrt{2^2 + 3^2}} = \frac{2}{\sqrt{13}}$.
Using Snell's law: $\sin \beta = \frac{\mu_1}{\mu_2} \sin \alpha = \frac{1}{1.5} \times \frac{2}{\sqrt{13}} = \frac{2}{1.5 \sqrt{13}} = \frac{4}{3\sqrt{13}}$.
The refracted vector is $\vec{OB} = C\hat{i} - 4\hat{j}$. The angle $\beta$ is with the vertical axis,so $\sin \beta = \frac{|C|}{\sqrt{C^2 + (-4)^2}} = \frac{C}{\sqrt{C^2 + 16}}$.
Equating the two expressions for $\sin \beta$: $\frac{C}{\sqrt{C^2 + 16}} = \frac{4}{3\sqrt{13}}$.
Squaring both sides: $\frac{C^2}{C^2 + 16} = \frac{16}{9 \times 13} = \frac{16}{117}$.
$117C^2 = 16(C^2 + 16) \implies 117C^2 = 16C^2 + 256$.
$101C^2 = 256 \implies C^2 = \frac{256}{101} \approx 2.534$.
$C = \sqrt{2.534} \approx 1.59 \approx 1.6$.