Question

An object is placed at a distance of 15 cm in front of a convex mirror of radius of curvature 10 cm. (a) Where will the image form ? (b) Find the magnification m. (c) What will be the nature of Image, real or virtual ?​

Answer 1

$\huge{ \underline{ \bf{Given \: : }}}$

• $\tt \: r = 10 \: cm$
• $\tt \: f = \frac{r}{2} = 5 \: cm \: (positive)$
• $\tt \: u = 15 \: cm \: (negative)$
• $\tt \: v = ?$

$\huge{ \underline{ \bf{To \: find \: : }}}$

• $\tt \: a. \: where \: the \: image \: will \: form$
• ${ \tt{b. \: magnification}} \: m$
• $\tt \: c. \: the \: nature \: of \: the \: image$

$\huge{ \underline{ \underline{ \mathrm{Solution \: : - }}}}$

A.

As we know,

From relation

$\purple \leadsto \star\red{ \underline{ \underline{ \boxed{ \tt{ \green{ \frac{1}{u} + \frac{1}{v} = \frac{1}{f} }}}}}} \star$

$\\ \tt \to \frac{1}{v} + \frac{1}{( - 15)} = \frac{1}{5}$

$\\ \tt \to \frac{1}{v} = \frac{1}{5} + \frac{1}{15}$

$\\ \tt \to \frac{1}{v} = \frac{3 + 1}{15}$

$\\ \tt \to \frac{1}{v} = \frac{4}{15}$

$\\ \tt \to \: v = \frac{15}{4}$

$\\ \tt \to \: v = 3.75 \: cm$

Thus the image will form at the a distance 3.75 cm behind the mirror.

B.

${ \tt \: magnification \: m }= - \frac{v}{u}$

$\tt = - \frac{3.75}{( - 15)}$

$\tt = \frac{1}{4}$

Thus the size of the image is one-fourth the size of the object.

C.

The image will be virtual and erect.