Question

A virtual image three times the size of the object is obtained with a concave mirror of radius of curvature 36cm. The distance of the object from the mirror is(a) 5cm(b) 12cm(c) 10cm(d) 20cm

Answer 1

Answer:

Explanation:

Hope it will help uhh

Answer 2

The distance from the mirror of object is 12 cm which produces a virtual image of 3 times of size of original object.

Answer: Option b

Solution:

Given:

$\begin{array}{l}{\text { Height of image }=3 \times \text { height of object }} \\ {\text { Radius of curvature of concave mirror }=36 \mathrm{cm}}\end{array}$

As we know that,

$\begin{array}{l}{\text {Magnification } m=\frac{\text {height of the image}}{\text {height of the object }}} \\ {\text {Magnification } m=\frac{\text {distance of image } v}{\text {distance of object } u}}\end{array}$

$\frac{\text {height of image}}{\text {height of object}}=\frac{-v}{u}$

Therefore,  

$\begin{array}{l}{-v=\frac{\text {height of image } \times u}{\text {height of object }}} \\ {-v=\frac{3 \text { height of object } \times u}{\text {height of object }}}\end{array}$

- v = 3 u

v = -3 u

Now, $R=2 \times focal\ length$

$f=\frac{R}{2}=18 \mathrm{cm}$

Len’s maker formula –

$\begin{array}{l}{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}} \\ {\frac{1}{18}=\frac{1}{-3 u}+\frac{1}{u}} \\ {\frac{1}{18}=\frac{2}{3 u}}\end{array}$

u = 12 cm