Question

An object of height 6 cm is placed on the principle
axis of a concave mirror of focal length fat a
distance of 4f. The height of image will be
(1) 2 cm
(2) 12 cm
(3) 4 cm
(4) 1.2 cm​

Answer 1

It is a concave mirror so it is negative

Height of the object=6cm

Focal length =f=-f

Radius of curvature is =2*f (f)

=2f

Distance of the object =4f (u) =-4f

That means it is beyond the centre of curvature

Mirror formula:1/v+1/u =1/f

Sub in formula

• 1/v+-1/4f=-1/f
• 1/v=-1/f-1/4f
• 1/v=-4+1/4f
• 1/v=-3/4f
• V=-4f/3

Height of the image = - (v/u) =hi/ho

• -1/3=x/6
• 3x=-6
• X=-2

Answer 2

Answer:

(1) 2 cm

Explanation:

Given: It is a concave mirror, so, it is negative.

Height of the object (ho) = 6cm

Focal length (f) = -f

Distance of the object (u) = - 4f

To find: The height of image (hi)

Solution:

Radius of curvature = 2 × focal length

Radius of curvature = 2f

Distance of the object (u) = - 4f

That means, it is beyond the centre of curvature.

Now, from mirror formula,

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

On substituting the values, we get

$\frac{1}{v } \: + \: ( - \frac{1}{4f} ) \: = \: - \frac{1}{f}$

$\frac{1}{v } = \: - \frac{1}{f} - \frac{1}{4f}$

$\frac{1}{v} \: = \: - \frac{3}{4f}$

$v \: = \: - \frac{4f}{3}$

Now, height of image will be calculated as follows:

$- \frac{v}{u} \: = \: \frac{hi}{ho}$

$- \frac{1}{3 } = \frac{x}{6}$

$3x \: = \: - 6$

$x \: = \: - 2$

Therefore, the height of image is 2cm.

Formula used:

• Mirror formula: $\frac{1}{v} \: + \frac{1}{u} \: = \: \frac{1}{f}$
• $- \frac{v}{u} \: = \: \frac{hi}{ho}$

The correct option for the above question is (1) 2cm.

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