Question

an object of height 5 cm is placed before a cancave mirror of focal length 20cm so that the real image is formed at the distance of 60cm from it what is the positve of object?

Answer 1

Explanation:

I t placed at the centre of curvature and it is placed at 30 cm away of the mirror

Answer 2

The height of the image (inverted) will be 10 cm

Explanation:

Given:

Height of the object = 5 cm

Focal length of the mirror f = 20 cm

Distance of image of the object  v = 60 cm

Since the mirror is concave and the image is real

Therefore, the image, object and the focal length will be on the same side of the mirror

Thus, all will have same sign

Using mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

$\implies \frac{1}{60}+\frac{1}{u}=\frac{1}{20}$

$\implies \frac{1}{u}=\frac{1}{20}-\frac{1}{60}$

$\implies \frac{1}{u}=\frac{3-1}{60}$

$\implies\frac{1}{u}=\frac{2}{60}$

$\implies\frac{1}{u}=\frac{1}{30}$

$\implies u=30}$ cm

We know that

Magnification

$m=-\frac{v}{u}=\frac{\text{Height of the image}}{\text{Height of the object}}$

$\implies \text{Height of the image}=-\frac{v}{u}\times\text{Height of the object}$

$\implies \text{Height of the image}=-\frac{60}{30}\times 5=-10$ cm

Negative sign indicates that the image will be inverted

Hope this answer is helpful.

Know More:

Q: An object of height 1.2m is placed before a concave mirror of focal length 20cm so that a real image is formed at a distance of 60cm from it. Find the position of an object. What will be the height of the image formed?

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