Solution:

Ans. (i) Refractive index of a medium is the ratio of speed of light (c) in free space to the speed of light (v) in that medium.

$$ \begin{align*} & \mu=\frac{c}{v} \ & \mu=\frac{c}{v}=\frac{1}{\sin i_{c}} \tag{ii}\ &=\frac{3 \times 10^{8}}{v}=\frac{1}{\frac{30}{50}} \end{align*} $$

$$ \begin{align*} v & =\frac{30}{50} \times 3 \times 10^{8} \ & =1.8 \times 10^{8} \mathrm{~m} / \mathrm{s} \end{align*} $$

[CBSE Marking Scheme, 2016]

[A] Q. 2. A ray $P Q$ incident normally on the refracting face $B A$ is refracted in the prism $B A C$ made of material of refractive index 1.5. Complete the path of ray through the prism. From which face will the ray emerge ? Justify your answer.

Ans.

At plane $A C$, the incident angle $=30^{\circ}$

If critical angle for total internal reflection $=S_{C}$

$$ \begin{aligned} n & =\frac{1}{\sin C} \ \Rightarrow \quad \sin C & =\frac{1}{1.5} \ & =0.67 \text { hence, } C>30^{\circ} \end{aligned} $$

$\left(\sin 30^{\circ}=0.5\right)$

As incident angle is less than critical angle, it would emerge out from $A C$. In the figure path of the ray is shown.

$$ \begin{array}{rlrl} & & \frac{\sin 30^{\circ}}{\sin e} & =\frac{1}{1.5} \ \therefore & \sin e & =1.5 \times \sin 30^{\circ} \ & \therefore & & =1.5 \times 0.5=0.75 \ & & e & =48^{\circ} \end{array} $$

Commonly Made Error

• Several students could not get angle $C$ correctly from $\frac{1}{\sin C}=n$ well understood.

[AI Q. 3. A convex lens is placed in contact with a plane mirror. A point object at a distance of $20 \mathrm{~cm}$ on the axis of this combination has its image coinciding with itself. What is the focal length of the lens?

[CBSE OD 2014]

Ans. According to the question

Object distance, $=20 \mathrm{~cm}$.

Image distance $=20 \mathrm{~cm}$

As the image $I$ of the object coincides with $O$, the rays refracted first from the lens and then reflected by the plane mirror must be retracing their path. It is the condition only when rays refracted by the convex lens fall normally on the mirror i.e, the refracted rays form a beam parallel to principal axis of the lens. Hence the object $O$ must be at the focus of the convex lens.

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[CBSE Marking Scheme 2014]