Question

The focal length of a concave lens is 15 cm .Find the nature and position of a image placed at a distance of 10cm from the lens.​

Answer 1

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☞Given u = -10cm

☞v= ?

☞ f = -15cm

✮Applying mirror's formula,

✮1/v +1/u = 1/f

1/v + 1/-10 = 1/-15

1/v = -1/15 + 1/10

=(-2+3)/30

= 1/30

V = 30 cm.

✮Hence, we get enlarged image, virtual in nature having magnification 3.

1/p+1/q=1/f

1/10 + 1/q = 1/15

1/q=2/30–3/30

q= -30

magnification(m)=-q/p

m= - -30/10= +3

⟹Image is virtual, erect, and 3 times larger