Question

If a $$3cm$$ tall object placed perpendicular to principal axis of a convex lens of focal length $$15cm$$ produces a real inverted image of height $$15cm$$, then its object distance $$(u)$$ is ……………. and image distance is …………….
A. $u=-18m,v=+90m$
B. $u=+18cm,v=-90cm$
C. $u=-18cm,v=+90cm$
D. $u=+18cm,v=+90cm$

Answer 1

Hint: Use the formula for magnification in terms of image and object heights. This will give you the value of $$m$$. Use this value to find the relation between $$u$$ and $$v$$. As the focal length is given, use the mirror formula and the relation between $$u$$ and $$v$$to get the required values. Use suitable sign conventions.

Complete step by step answer:
It is given that the height of the image (h2) = 15 cm
As the image formed is inverted, therefore, h2 = -15 cm
It is given that the height of the object (h1) is 3 cm.
We know that the magnification of a mirror is defined as the ratio of the height of an image to the height of an object. Therefore, magnification of the lens is given by,
$$m=\left|{\displaystyle \frac{h_2}{h_1}}\right|$$
$$\begin{gathered} \Rightarrow m=\left|{\displaystyle \frac{-15}{3}}\right| \\ \Rightarrow m=5 \end{gathered}$$

Magnification of a mirror is also equal to the ratio of image distance to that of object distance. Therefore, $$m={\displaystyle \frac{v}{u}}$$, where, u is the object distance and v is the image distance.
$$\begin{gathered} 5={\displaystyle \frac{-v}{-u}} \\ \Rightarrow {\displaystyle \frac{-v}{-u}}=5 \\ \Rightarrow v=5u \end{gathered}$$
It is given that focal length of the lens, f = 15 cm
We know that the ${\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}$ ,
where f is the focal length of the mirror
$$\begin{gathered} \Rightarrow {\displaystyle \frac{1}{15}}={\displaystyle \frac{1}{v}}-\left({\displaystyle \frac{1}{-u}}\right) \\ \Rightarrow {\displaystyle \frac{1}{15}}={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}} \end{gathered}$$
Putting, $v=5u$ in the above equation, we get
$$\begin{gathered} {\displaystyle \frac{1}{15}}={\displaystyle \frac{1}{5u}}+{\displaystyle \frac{1}{u}} \\ \Rightarrow {\displaystyle \frac{1+5}{5u}}={\displaystyle \frac{1}{15}} \\ \Rightarrow u=18 \end{gathered}$$
Putting $u=18$ in $v=5u$ we get $v=90$
Therefore, the image will be formed at a distance 0f 90 cm from the mirror.

Hence, the correct option is D.

Note: The distance on the left hand side of the principal axis is taken as positive and on the right hand side of the principal axis is taken as negative. The height of the object is always taken as positive and the height of the inverted image is taken as negative.