Question

Find the focal length and nature of a lens which should be placed in contact with the lens of focal length 10 cm so that the power of combination becomes 5 Diopter

Answer 1

Power(P2) of lens with 10 cm focal length = 10010=10D
Let Power of the lens is P1
Total power of combination = P1 +P2
​5D=​ P1 + 10D
P1 = -5D
focal length = −15=−0.2m⇒−20 cm
Power of the lens is negative so the given lens is concave lens

Answer 2

Answer:

Power of the lens exists negative so the provided lens stands concave lens.

Explanation:

The length between the convex lens or a concave mirror and the focal point of a lens or mirror exists named the focal length. It exists at the point where parallel rays of light meet or converge.

Here, $f_{1}=?$

$f_{2}=10 \mathrm{~cm}$

$P_{2}=\frac{100}{f_{2}}=\frac{100}{10}=10D$

$P=5 D .$

As $P_{1}+P_{2}=P$

$P_{1}+10=5$

ог

$P_{1}==5-10=-5 D$. and

$f_{1}=\frac{100}{P_{1}}$

$=\frac{100}{-5}$

Thus,

$=-20 \mathrm{~cm}$

The lens must be concave.

Power of the lens exists negative so the provided lens stands concave lens.

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