Solution:

Ans. Using mirror formula

$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$

For plane mirror $f=\mathrm{\infty }$

Hence

$$\begin{array}{}\text{(1)}& \frac{1}{v}+\frac{1}{u}=0\end{array}$$

So, $v=-u$

It means the image sat equal distance and in opposite side of object. This is the true condition in image formation through plane mirror. Hence spherical mirror formula holds equally to a plane mirror. $\mathbf{1}$

Long Answer Type Questions

AI Q. 1.(i) Draw a ray diagram to show image formation when the concave mirror produces a real, inverted and magnified image of the object.

(ii) Obtain the mirror formula and write the expression for the linear magnification.

(iii) Explain two advantages of a reflecting telescope over a refracting telescope.

A [Delhi& O.D. 2018]

Ans. (i) Ray diagram to show the required image formation

(ii) Derivation of mirror formula Expression for linear magnification

iii) Two advantages of a reflecting telescope over a refracting telescope

$1/2+1/2$

(i)

(ii) In the above figure $\mathrm{△}BAP$ and $\mathrm{△}{B}^{\prime }{A}^{\prime }P$ are similar

$$\begin{array}{}\text{(i)}& \Rightarrow \frac{BA}{B^{\prime }{A}^{\prime }}=\frac{PA}{P{A}^{\prime }}\end{array}$$

Similarly, $\mathrm{△}MNF$ and $\mathrm{△}{B}^{\prime }{A}^{\prime }F$ are similar

$$\begin{array}{}\text{(ii)}& \Rightarrow \frac{MN}{B^{\prime }{A}^{\prime }}=\frac{NF}{F{A}^{\prime }}\end{array}$$

$$\begin{array}{rlrl}MN& =BANF& \approx PFF{A}^{\prime }& =P{A}^{\prime }-PF\end{array}$$

$\therefore$ Equation (ii) takes the following form

$$\begin{array}{}\text{(iii)}& \frac{BA}{B^{\prime }{A}^{\prime }}=\frac{PF}{P{A}^{\prime }-PF}\end{array}$$

Using equation (i) and (iii)

$$\frac{PA}{P{A}^{\prime }}=\frac{PF}{P{A}^{\prime }-PF}$$

For the given figure, as per the sign convention,

$$\begin{array}{rlrlrlr}PA& =-uP{A}^{\prime }& =-vPF& =-f\Rightarrow \frac{-u}{-v}& =\frac{-f}{-v-(-f)}\frac{u}{v}& =\frac{f}{v-f}uv-uf& =vf\end{array}$$

Dividing each term by $uvf$, we get

$$\begin{array}{rlr}\frac{1}{f}-\frac{1}{v}& =\frac{1}{u}\Rightarrow \frac{1}{f}& =\frac{1}{v}+\frac{1}{u}\end{array}$$

Linear magnification $=-\frac{-v}{u}$,

$$\text{ (alternatively, }m=\frac{h_i}{h_0}\text{ ) }1/2$$

(iii) Advantages of reflecting telescope over refracting telescope

(i) Mechanical support is easier

$1/2+1/2$

(ii) Magnifying power is large

(iii) Resolving power is large

(iv) Spherical aberration is reduced

(v) Free from chromatic aberration

(any two)

[CBSE Marking Scheme 2018]