Question

An object is placed at a distance 20cm in front of a convex mirror of radius of curvature 30cm. find the position and nature of the image

Answer 1

Hey,,
distance of the object (u) = 20 cm
radius = 30 cm
position of image(u)=?
Now,,
focal length (f) = 1/2r
=1/2*30
f=15cm.
Applying formula,
1/f=1/u+1/v
1/15=1/-20+1/v
1/15=v-20/-20v
-5v=300
v=-60cm.
Nature of the image is erect, virtual and diminished image.

Hope this helps!!

Answer 2

Given :-

A convex mirror has a radius of 20 cm. An object is placed 30 cm in front of the mirror.

To Find :-

What is the distance where the image is formed.

Formula Used :-

$\clubsuit$ ♣ Focal Length Formula :

$\begin{lgathered}\longmapsto \sf\boxed{\bold{\pink{f =\: \dfrac{Radius}{2}}}}\\\end{lgathered}$

where,

f = Focal Length

$\clubsuit$ ♣ Mirror Formula :

$\begin{lgathered}\longmapsto \sf\boxed{\bold{\pink{\dfrac{1}{v} + \dfrac{1}{u} =\: \dfrac{1}{f}}}}\\\end{lgathered}$

where,

v = Image Distance

u = Object Distance

f = Focal Length

Solution :-

First, we have to find the focal length :

Given :

Radius = 20 cm

According to the question by using the formula we get,

$\implies \sf f =\: \dfrac{\cancel{20}}{\cancel{2}}$

$\implies \sf f =\: \dfrac{\cancel{10}}{\cancel{1}}$

$\implies \sf\bold{\purple{f =\: 10\: cm}}$

Hence, the focal length is 10 cm .

Now, we have to find the image distance :

Given :

$\bigstar$ Focal length = 10 cm

$\bigstar$ Object Distance = - 30 cm

According to the question by using the formula we get,

$\begin{lgathered}\sf\bold{\green{\bigstar\: \: \: \dfrac{1}{v} =\: \dfrac{1}{f} - \dfrac{1}{u}\: \: \: \bigstar}}\\\end{lgathered}$

$\begin{lgathered}\longrightarrow \sf \dfrac{1}{v} =\: \dfrac{1}{10} - \bigg(- \dfrac{1}{30}\bigg)\\\end{lgathered}$

$\begin{lgathered}\longrightarrow \sf \dfrac{1}{v} =\: \dfrac{1}{10} + \dfrac{1}{30}\\\end{lgathered}$

$\begin{lgathered}\longrightarrow \sf \dfrac{1}{v} =\: \dfrac{3 + 1}{30}\\\end{lgathered}$

$\longrightarrow \sf \dfrac{1}{v} =\: \dfrac{4}{30}$

By doing cross multiplication we get,

$\longrightarrow \sf 4v =\: 30(1)$

$\longrightarrow \sf 4v =\: 30 \times 1$

$\longrightarrow \sf 4v =\: 30$

$\longrightarrow \sf v =\: \dfrac{30}{4}$

$\longrightarrow \sf\bold{\red{v =\: 7.5\: cm}}$

$\therefore$ The distance where the image is formed is 7.5 cm.