Question

the image of an object formed by a mirror is real inverted and is of magnification -1 if the image is at the distance of 30 cm from mirror, where is the object placed ? find the position of the image if the object is now moved 20cm towards the mirror what is the nature of the image obtained ? justify your answer with the help of ray diagram .

Answer 1

Solution is here if you like please give thanks

Answer 2

Answer:The image is virtual and enlarge in size 3 times

Explanation:

Given that,

Magnification m = -1

The distance of the image v = -30 cm

We know that,

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

$-1 = \dfrac{30}{u}$

$u = -30 cm$

So, the object is placed at 30 cm from the mirror

Now ,Using mirror's formula

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

$\dfrac{1}{f} = \dfrac{1}{-30} +\dfrac{1}{-30}$

$\dfrac{1}{f}= \dfrac{-2}{30}$

$f = -15 cm$

If the object is moved 20 cm towards the mirror

then, u = -10

$\dfrac{1}{-15} = \dfrac{1}{v}+\dfrac{1}{-10}$

$\dfrac{1}{v} = \dfrac{1}{30}$

$v = 30 cm$

So, the position of the image is  30 cm at back of the mirror

Now, magnification is

$m = \dfrac{-v}{u}$

$m =- \dfrac{30}{-10}$

$m = 3$

Hence, the image is virtual and enlarge in size 3 times.