Question

A mirror forms an erect image 40cm from the object and one third its height where must the mirror be situated ​

Answer 1

The mirror is located 30 cm from the object

Explanation:

To solve the problem, we can use the magnification equation, which states that:

$M=\frac{y'}{y}=-\frac{q}{p}$

where

M is the magnification

y' is the size of the image

y is the size of the object

q is the distance of the image  from the mirror

p is the distance of the object from the mirror

In this problem, we notice that the image formed by the mirror is erect and diminished: this means that this is a convex mirror, so the image is virtual, and this means that the image and the object are located on opposite sides of the mirror. Therefore,

$p-q=40 cm$ (distance of the image from the object is 40 cm, but since the image is virtual, q is the negative)

The size of the image is 1/3 that of the object, so

$y'=\frac{1}{3}y$

Solving the equation for p, we find the distance between the object and the mirror:

$\frac{y'}{y}=-\frac{p-40}{p}\\\\p=\frac{40}{1+\frac{y'}{y}}=\frac{40}{4/3}=30 cm$

So, the mirror is 30 cm from the object.

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