Question

consider the situation where an object is 3cm height mirror is concave with 6 cm focal length object is placed at the centre of curvature​

Answer 1

Answer:

ans is -1 magnification

Answer 2

Given:

Object = 3 cm (height).

Focal length(f)=6cm.  

The magnification produced by the mirror=?

Explanation:

The centre of curvature(u) of a mirror is twice its focal length.

$u=2*f$

$u=2*6$

$u=12cm$.

To find the distance of the image(v) from the mirror,formula is

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

v is the distance of the image,

u is the distance of the object from the mirror and

f is the focal length.      

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

$\frac{1}{v} =\frac{1}{6}- \frac{1}{12}$

$\frac{1}{v}= \frac{1}{12}$

$v=12cm$

The linear magnification produced by a spherical lens is the ratio of the image size to the object size formed by the lens  measured perpendicular to the principal axis.

$m=\frac{v}{u}$

$m=\frac{12}{12}$

$m=1$

The mirror will produce an image of magnification (m)=1.