If $\hat{i}$ denotes a unit vector along the incident light ray,$\hat{r}$ a unit vector along the refracted ray into a medium of refractive index $\mu$,and $\hat{n}$ a unit vector normal to the boundary of the medium directed towards the incident medium,then the law of refraction is:

  • A
    $\hat{i} \cdot \hat{n} = \mu (\hat{r} \cdot \hat{n})$
  • B
    $\hat{i} \times \hat{n} = \mu (\hat{n} \times \hat{r})$
  • C
    $\hat{i} \times \hat{n} = \mu (\hat{r} \times \hat{n})$
  • D
    $\mu (\hat{i} \times \hat{n}) = \hat{r} \times \hat{n}$

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Similar Questions

The $x-z$ plane separates two media $A$ and $B$ with refractive indices $\mu_1$ and $\mu_2$ respectively. $A$ ray of light travels from $A$ to $B$. Its directions in the two media are given by the unit vectors $\vec{r}_A = a\hat{i} + b\hat{j}$ and $\vec{r}_B = \alpha\hat{i} + \beta\hat{j}$ respectively,where $\hat{i}$ and $\hat{j}$ are unit vectors in the $x$ and $y$ directions. Then:

An opaque sphere of radius $a$ is just immersed in a transparent liquid as shown in the figure. $A$ point source is placed on the vertical diameter of the sphere at a distance $a/2$ from the top of the sphere. One ray originating from the point source after refraction from the air-liquid interface forms a tangent to the sphere. The angle of refraction for that particular ray is $30^{\circ}$. The refractive index of the liquid is:

$A$ transparent slab of thickness $d$ has a refractive index $n(z)$ that increases with $z$. Here $z$ is the vertical distance inside the slab,measured from the top. The slab is placed between two media with uniform refractive indices $n_1$ and $n_2 (> n_1)$,as shown in the figure. $A$ ray of light is incident with angle $\theta_i$ from medium $1$ and emerges in medium $2$ with refraction angle $\theta_f$ with a lateral displacement $l$. Which of the following statement$(s)$ is(are) true?
$(A)$ $n_1 \sin \theta_i = n_2 \sin \theta_f$
$(B)$ $n_1 \sin \theta_i = (n_2 - n_1) \sin \theta_f$
$(C)$ $l$ is independent of $n_2$
$(D)$ $l$ is dependent on $n(z)$

Immiscible transparent liquids $A, B, C, D$ and $E$ are placed in a rectangular glass container,forming layers based on their densities. The refractive indices of the liquids are given in the table below. The container is illuminated from the side,and a small piece of glass with a refractive index of $1.61$ is gently dropped into the liquid layers. In which liquid will the glass piece not be visible as it descends?
| Liquid | Refractive Index |
| :--- | :--- |
| $A$ | $1.51$ |
| $B$ | $1.53$ |
| $C$ | $1.61$ |
| $D$ | $1.52$ |
| $E$ | $1.65$ |

$A$ light beam is travelling 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:

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