Question

Answer the above attachment


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Answer 1

ANSWER :

Given : h1=4.5cm

u=−12cm

f=15cm

(1/u)+ (1/v) =(1/f)

Therefore, (1/v) = (1/f)−(1/u)

= (1/15) + (1/12)

= (4+5)/ (60)

= (9/60)

Therefore v= (60/9) =6.7cm

The image is virtual, formed at 6.7 cm, at the back of the mirror.

m= (h2/h1)

=− (v/u)

(h2/4.5)=−(6.7/12)

h2= (6.7×4.5)/(12)

=2.5cm

The image is erect and virtual. Size of image is 2.5 cm and its position is 6.67cm behind the mirror. So, when the needle is moved farther from the mirror, the size of image decreases.

ANSWER BY ARMAN