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Light: Reflection and Refraction - Explain image formation by lenses, lens formula, magnification, and power of a lens

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The power of a lens is defined as the measure of the degree of convergence or divergence of light rays falling on it.

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Mathematically, the power of a lens is the reciprocal of its focal length ff.

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The SI unit of power of a lens is Dioptre, represented by the symbol DD.

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One Dioptre (1D1 D) is the power of a lens whose focal length is 1 m1 \text{ m}. Thus, 1D=1 m−11 D = 1 \text{ m}^{-1}.

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The power of a convex (converging) lens is positive (++) because its focal length is positive.

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The power of a concave (diverging) lens is negative (−-) because its focal length is negative.

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When multiple thin lenses are placed in contact, the net power of the combination is the algebraic sum of the individual powers of the lenses.

📐Formulae

P=1f(in meters)P = \frac{1}{f (\text{in meters})}

P=100f(in centimeters)P = \frac{100}{f (\text{in centimeters})}

Pnet=P1+P2+P3+⋯+PnP_{net} = P_1 + P_2 + P_3 + \dots + P_n

💡Examples

Problem 1:

A convex lens has a focal length of 20 cm20 \text{ cm}. Calculate its power.

Solution:

Given: f=+20 cm=+0.2 mf = +20 \text{ cm} = +0.2 \text{ m}. Using the formula P=1fP = \frac{1}{f}, P=10.2=+5DP = \frac{1}{0.2} = +5 D.

Explanation:

Since the lens is convex, the focal length is taken as positive. The resulting power is +5D+5 D, indicating a converging lens.

Problem 2:

A person uses a lens of power −2.5D-2.5 D for correcting their vision. Find the focal length and type of the lens.

Solution:

Given: P=−2.5DP = -2.5 D. Using the formula f=1Pf = \frac{1}{P}, f=1−2.5 m=−0.4 m=−40 cmf = \frac{1}{-2.5} \text{ m} = -0.4 \text{ m} = -40 \text{ cm}.

Explanation:

The negative sign of the focal length and power indicates that the lens is a concave (diverging) lens.

Problem 3:

Two lenses of powers +3.5D+3.5 D and −1.5D-1.5 D are placed in contact. Find the net power and the net focal length of the combination.

Solution:

Net Power P=P1+P2P = P_1 + P_2 P=(+3.5D)+(−1.5D)=+2.0DP = (+3.5 D) + (-1.5 D) = +2.0 D. Net focal length f=1P=12.0 m=+0.5 m=+50 cmf = \frac{1}{P} = \frac{1}{2.0} \text{ m} = +0.5 \text{ m} = +50 \text{ cm}.

Explanation:

The powers are added algebraically. The positive net power indicates that the combination behaves like a convex lens.