Question

at what distance from convex mirror of focal length 2.5 cm should a boy stand so that this image has a hight equal to the half of the original height

Answer 1

Heya

Magnification=height of the image/height of the object = -image distance/object distance

Let the actual height be x

=(x/2)/x

=1/2

1/2= -v/u

Using sign convention,

1/2 = v/u

u=2v

That is object distance is twice the image distance

So, the object should be placed at the centre of curvature and the image will be formed at focus.

Object should be placed at centre of curvature =2(2.5)=5cm

The image will be formed at focus =2.5cm

Answer 2

Answer:

The boy must stand at a distance of 2.5 cm from a convex lens so that this image has a hight equal to the half of the original height.

Explanation:

Let us assume that the height of the boy is $x$

Then, as per the given information, the height of the image should be $\frac{x}{2}$

Given focal length of convex mirror is 2.5 cm.

Magnification equation states that,  $M=\frac{hi}{ho}=-\frac{di}{do}$

where, $M$ is magnification, $hi$ is image height, $ho$ is object height,

$di$ is image distance, and $do$ is object distance.

Applying the equation  to the given data, we get

$M= \frac{1}{2}=-\frac{di}{do}$

∴ $do=-2di$

Now the mirror equation states that, $\frac{1}{f}=\frac{1}{do}=\frac{1}{di}$

Applying the equation to the given data, we get

$\frac{1}{2.5}=\frac{1}{do}=\frac{-1}{2do}$

∴ $do$ = 2.5 cm