Question

$\huge{\purple\bigstar\underline{\bf{Questions:-}}}$
1. Express the power (with sign) of a concave lens of focal length length 20 cm.
2. The power of a lens is -2.0D. Find its focal length.

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Answer should be explained!

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Answer 1

Answer:

1. Power = -5 D
2. Focal length = -0.5 m

Explanation:

Concepts used:

1. First, let's see the sign convention:

• Power for a concave lens is always negative and
• Power for a convex lens is always positive.

2. Power of lens = 1 ÷ Focal length

Solution for Question 1:

Given:

• Concave lens
• Focal length = 20 cm = 0.2 m

To find:

Power = ?

Solution:

$\bf{\pink{Power = \dfrac{1}{Focal \: \: Length}}}$

$\implies \sf{Power = \dfrac{1}{0.2}}$

$\implies \sf{Power = 5\:D}$

It is given that the lens is concave.

$\therefore \: \boxed{\red{\boldsymbol{Power = -5 \:D}}}$

Solution for Question 2:

Given:

• Power = -2 D

To find:

Focal length = ?

Solution:

$\green{\bf{Power = \dfrac{1}{Focal \: length}}}$

$\implies \sf{-2=\dfrac{1}{Focal \: length}}$

$\implies \sf{-2 \times (Focal \: length)=1}$

$\implies \sf{Focal \: length = \dfrac{-1}{2}}$

$\therefore \: \boxed{\blue{\boldsymbol{Focal \: \: Length = -0.5 \:m}}}$

Answer 2

Answer:

1. Power of lens = -5 D
2. Focal length of lens = -0.5 m

Step-by-step explanation:

We know that,

1•} In the laws of sign convention :-

• Power of a concave lens is always negative
• Power of a convex lens is always positive.

2•} Power of lens = $\sf\dfrac{1}{Focal \: Length(in \: metres)}$

Solution 1 :-

❍ Given :-

• Concave lens
• Focal length = 20 cm = 0.2 m

❍ To Find :-

• The power of lens

❍ Solution :-

Using the formula,

$\fbox{\sf→Power = \dfrac{1}{Focal \: Length(in \: metres)}}$

$\sf→Power = \dfrac{1}{0.2}$

$\sf→\fbox\green{Power = 5D}$

Hence, the lens is concave. Therefore, the power of lens is -5D.

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Solution 2 :-

❍ Given :-

• Power of lens = -2.0D

❍ To Find :-

• Focal Length of lens

❍ Solution :-

We know that,

$\fbox{\sf→Power = \dfrac{1}{Focal \: Length(in \: metres)}}$

$\sf→ - 2.0D = \dfrac{1}{Focal \: Length(in \: metres)}$

$\sf→ - 2 \times (Focal \: Length) = 1$

$\sf→ Focal \: Length = \dfrac{1}{ - 2}$

$→\fbox\green{\sf Focal \: Length = - 0.5m}$

Hence, the focal length of given lens is -0.5 m.