A light ray incident along a vector vec{AO} ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$1.6$

Vedclass · Answer

(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, $	an \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: $\sin \beta = \frac{\mu_1}{\mu_2} \sin \alpha = \frac{1}{1.5} 	imes \frac{2}{\sqrt{13}} = \frac{2}{1.5 \sqrt{13}} = \frac{4}{3\sqrt{13}}$.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: $\frac{C^2}{C^2 + 16} = \frac{16}{9 	imes 13} = \frac{16}{117}$.$117C^2 = 16(C^2 + 16) \implies 117C^2 = 16C^2 + 256$.$101C^2 = 256 \implies C^2 = \frac{256}{101} \approx 2.534$.$C = \sqrt{2.534} \approx 1.59 \approx 1.6$.

$A$ light ray incident along a vector $\vec{AO}$ $(\vec{AO} = 2\hat{i} - 3\hat{j})$ emerges along a vector $\vec{OB}$ $(\vec{OB} = C\hat{i} - 4\hat{j})$ as shown in the figure. The value of $C$ is . . . . . . .

Similar Questions

Light of wavelength $300 \ nm$ in medium $A$ enters into medium $B$ through a plane surface. If the frequency of light is $5 \times 10^{14} \ Hz$ and the ratio of speed in medium $A$ to that in medium $B$ is $\frac{4}{5}$,the absolute refractive index of medium $B$ is:

The refractive index of glass is $1.6$ and the speed of light in glass will be . . . . . . . Speed of light in vacuum is $3 \times 10^{8} \,m \,s^{-1}$.

$A$ fish in water (refractive index $n$) looks at a bird vertically above in the air. If $y$ is the height of the bird and $x$ is the depth of the fish from the surface,then the distance of the bird as estimated by the fish is

$A$ light beam is traveling from Region $I$ to Region $IV$ (Refer Figure). The refractive indices in Regions $I, II, III$ and $IV$ are $n_0, \frac{n_0}{2}, \frac{n_0}{6}$ and $\frac{n_0}{8}$,respectively. The angle of incidence $\theta$ for which the beam just misses entering Region $IV$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module