Question

focal length of diverging mirror is 15 cm. An object is kept at a distance of 30 cm from the mirror. Find the image distance and its nature​

Answer 1

Answer:

Image distance = 10 cm

Nature = Virtual and erect

Explanation:

(Diverging mirror is also called convex mirror.)

As per the provided information in the given question, we have :

• Focal length (f) = +15 cm (As it is a convex mirror)
• Object distance (u) = – 30 cm (as it is placed left to the mirror)

We are asked to calculate the image distance and its nature.

In order to find the image distance, we need to apply the mirror formula.

$\longrightarrow \sf{\quad { \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}$

• v denotes image distance
• u denotes object distance
• f denotes focal length

$\longrightarrow \sf{\quad { \dfrac{1}{v} + \dfrac{1}{-30} = \dfrac{1}{15} }}$

$\longrightarrow \sf{\quad { \dfrac{1}{v} - \dfrac{1}{30} = \dfrac{1}{15} }}$

$\longrightarrow \sf{\quad { \dfrac{1}{v}= \dfrac{1}{15} + \dfrac{1}{30} }}$

$\longrightarrow \sf{\quad { \dfrac{1}{v}= \dfrac{2 + 1}{30} }}$

$\longrightarrow \sf{\quad { \dfrac{1}{v}= \dfrac{3}{30} }}$

On reciprocating both sides,

$\longrightarrow \sf{\quad { v= \dfrac{30}{3} }}$

$\longrightarrow \quad \underline{\boxed{ \textbf{\textsf{ v = 10 \; cm }}}}$

Therefore, image distance is 10 cm.

The image is formed behind the mirror, so the nature of the image is erect and virtual.