Question

A compound lens is formed with two lenses if powers +15.5D and -5.5D kept in contact.A 3cm long object is placed at distance of 30cm from the lens.What is the length of the image?

Answer 1

The net power is P1 + P2 = 15.5 - 5.5 D = 10.0 D 
1/f = P(D)
1/f = 10
⇒f = 1/10 = 0.1 m = 10 cm
⇒f = 10 cm

A/Q
h = 3 cm
u = -30 cm
f = 10 cm

According to the Lens formula : 1/v=1/u+1/f
⇒ 1/v = - 1/30 + 1/10 (3/3)      <By LCM>
⇒1/v = -1+3/30
⇒i/v = 2/30
⇒1/v = 1/15
⇒v = 15 cm.
image height / h = v/u
image height = 15 * 3/-30
⇒ = -1.5 cm

Hence the image distance is 15 cm and the size of the image is 1.5 cm below the principal Axis. 

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Yours, Jahnavi

Answer 2

The net power is P1 + P2 = 15.5 - 5.5 D = 10.0 D

1/f = P(D) 1/f = 10⇒f = 1/10 = 0.1 m = 10 cm⇒f = 10 cm

A/Q

h = 3 cmu = -30 cmf = 10 cm

According to the Lens formula : 1/v=1/u+1/f⇒ 1/v = - 1/30 + 1/10 (3/3)

<By LCM>

A/Qh = 3 cmu = -30 cmf = 10 cmAccording to the Lens formula : 1/v=1/u+1/f⇒ 1/v = - 1/30 + 1/10 (3/3) <By LCM>⇒1/v = -1+3/30⇒i/v = 2/30⇒1/v = 1/15⇒v = 15 cm.

.image height / h = v/u

image height = 15 * 3/-30 = -1.5 cm

Hence the image distance is 15 cm and the size of the image is 1.5 cm below the principal Axis.

. Thanks........Mark as brainliest if helpful.......