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Selina solutions

CHAPTERS

1. Chapter 1 – Measurements and Experimentation

2. Chapter 2 Motion In One Dimension

4. Chapter 4 Pressure In Fluids And Atmospheric Pressure

5. Chapter 5 Upthrust In Fluids Archimedes Principle And Floatation

7. Chapter 7 Reflection Of Light

8. Chapter 8 Propogation Of Sound Waves

**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)**

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

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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