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Chapter 7 Reflection Of Light | Exercise 7C

Question 48

A concave mirror forms a real image of an object placed in front of it at a distance 30cm, of size three times the size of the object. Find (a) the focal length of the mirror (b)

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Given:

(a)Distance of the object (u) = 30cm (negative)

We know that,

\mathrm{m}=\frac{1}{0}=\frac{30}{0}=3

For a real object, the value of ‘m’ is negative

m = -3

Asm=-\frac{v}{u}

=> v = 3u

=> v = 3 x 30 = 90 cm

Therefore, the image is formed 90cm in front of the mirror.

Mirror formula can be given by

Therefore, the image is formed 90cm in front of the mirror.

The mirror formula can be given by

\begin{array}{c} \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \frac{1}{f}=\frac{1}{30}+\frac{1}{90}=-\frac{4}{90} \\ f=-22.5 \mathrm{cm} \end{array}

The focal length of the mirror is 22.5cm

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Our top 5% students will be awarded a special scholarship to Lido.

subject-cta

Question 48

A concave mirror forms a real image of an object placed in front of it at a distance 30cm, of size three times the size of the object. Find (a) the focal length of the mirror (b)

Looking to do well in your science exam ? Learn from an expert tutor. Book a free class!

Given:

(a)Distance of the object (u) = 30cm (negative)

We know that,

\mathrm{m}=\frac{1}{0}=\frac{30}{0}=3

For a real object, the value of ‘m’ is negative

m = -3

Asm=-\frac{v}{u}

=> v = 3u

=> v = 3 x 30 = 90 cm

Therefore, the image is formed 90cm in front of the mirror.

Mirror formula can be given by

Therefore, the image is formed 90cm in front of the mirror.

The mirror formula can be given by

\begin{array}{c} \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \frac{1}{f}=\frac{1}{30}+\frac{1}{90}=-\frac{4}{90} \\ f=-22.5 \mathrm{cm} \end{array}

The focal length of the mirror is 22.5cm

Our top 5% students will be awarded a special scholarship to Lido.

subject-cta
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