Question

An object is placed at a distance of 15 cm in front of a concave mirror of focal length 12 cm. Find the position and magnificent of the image. ​

Answer 1

AnsWer :

-60 Cm.

Given :

• (u) Object distance -15 cm.
• (f) focal length -12 cm.
• (v) image distance ?

Formula Use :

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

Where,

• (f) focal length.
• (v) Image distance.
• (u) Object distance.

$\tt M = \frac{h_i}{h_o} = \frac{ - v}{u}$

• (M) magnified.
• (hi) height of image.
• (ho) height of object.

Solution :

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

$\implies \tt \frac{1}{ - 12} = \frac{1}{v} + \frac{1}{ - 15}$

$\implies \tt \frac{1}{ 15} - \frac{1}{ 12} = \frac{1}{v}$

$\implies \tt \frac{4 - 5}{60} = \frac{1}{v}$

$\implies \tt \frac{1}{v} = \frac{ - 1}{60}$

$\implies \tt v = - 60 \: cm.$

Now,

$\tt M = \frac{h_i}{h_o} = \frac{ - v}{u}$

$\tt \implies \frac{ - v}{u} = \frac{ - ( - 60)}{ - 15}$

$\tt = - 4 \: cm.$

Therefore, the when we place image at distance 60 cm it gives Sharp image and magnitude of image be 4 cm.

Answer 2

Answer:

The image will be formed 60 cm in front of the mirror.

The magnificent of the image will be -4.

Step-by-step explanation:

Referred to the attachments.