Question

A convex mirror has a radius of 20cm .An object is placed 30cm in front of the mirror. Find the distance where the image is formed​

Answer 1

Answer :-

The image is formed at a distance of 7.5 cm from the mirror .

Explanation :-

We have :-

→ Radius of curvature = 20 cm

→ Distance of the object = 30 cm

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Firstly, let's calculate the focal length of the mirror.

⇒ f = R/2

⇒ f = 20/2

⇒ f = 10 cm

Since, the mirror is convex :-

• f = + 10 cm

• u = - 30 cm

According to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = 1/10 - 1/(-30)

⇒ 1/v = 1/10 - (-1)/30

⇒ 1/v = (3 + 1)/30

⇒ 1/v = 4/30

⇒ 4v = 30

⇒ v = 30/4

⇒ v = 7.5 cm

Answer 2

Answer:

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 :

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

where,

• f = Focal Length

$\clubsuit$ Mirror Formula :

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

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,

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

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

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

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

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

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.