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Ray Optics and Optical Instruments - Lens Maker’s Formula

Grade 12CBSEPhysics

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

🔑Concepts

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The Lens Maker’s Formula relates the focal length ff of a lens to the refractive index nn of its material and the radii of curvature R1R_1 and R2R_2 of its two surfaces.

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It is derived based on the assumptions that the lens is 'thin' (thickness is negligible) and the rays are paraxial (small angles with the principal axis).

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Sign Convention is critical: Distances measured in the direction of incident light are positive. For a biconvex lens, R1R_1 is usually positive and R2R_2 is negative.

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The focal length depends on the surrounding medium. If a lens of refractive index n2n_2 is placed in a medium of refractive index n1n_1, the term (n−1)(n - 1) becomes (n2n1−1)(\frac{n_2}{n_1} - 1).

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A lens may change its nature (from converging to diverging) if immersed in a medium with a refractive index greater than the material of the lens (nmedium>nlensn_{medium} > n_{lens}).

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For a plano-convex lens, one of the radii of curvature is infinity (R=∞R = \infty), which makes 1R=0\frac{1}{R} = 0.

📐Formulae

1f=(n21−1)(1R1−1R2)\frac{1}{f} = (n_{21} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

1f=(n2n1−1)(1R1−1R2)\frac{1}{f} = \left( \frac{n_2}{n_1} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

P=1fP = \frac{1}{f}

n2v−n1u=n2−n1R\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}

💡Examples

Problem 1:

A biconvex lens has radii of curvature 20 cm20\text{ cm} and 30 cm30\text{ cm}. The refractive index of the glass is 1.51.5. Calculate its focal length in air.

Solution:

Given: R1=+20 cmR_1 = +20\text{ cm}, R2=−30 cmR_2 = -30\text{ cm} (by sign convention), and n=1.5n = 1.5. Using Lens Maker's Formula: 1f=(n−1)(1R1−1R2)\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). Substituting values: 1f=(1.5−1)(120−1−30)=0.5(3+260)=0.5×560=124\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-30} \right) = 0.5 \left( \frac{3 + 2}{60} \right) = 0.5 \times \frac{5}{60} = \frac{1}{24}. Thus, f=24 cmf = 24\text{ cm}.

Explanation:

The focal length is positive, indicating that the biconvex lens acts as a converging lens in air.

Problem 2:

A glass lens (ng=1.5n_g = 1.5) has a focal length of 20 cm20\text{ cm} in air. Find its focal length when immersed in water (nw=1.33n_w = 1.33).

Solution:

In air: 1fa=(ng−1)K\frac{1}{f_a} = (n_g - 1) K where K=(1R1−1R2)K = (\frac{1}{R_1} - \frac{1}{R_2}). 120=(1.5−1)K⇒K=110\frac{1}{20} = (1.5 - 1) K \Rightarrow K = \frac{1}{10}. In water: 1fw=(ngnw−1)K\frac{1}{f_w} = (\frac{n_g}{n_w} - 1) K. Substituting KK: 1fw=(1.51.33−1)×110≈(1.127−1)×0.1=0.0127\frac{1}{f_w} = (\frac{1.5}{1.33} - 1) \times \frac{1}{10} \approx (1.127 - 1) \times 0.1 = 0.0127. Thus, fw≈10.0127≈78.74 cmf_w \approx \frac{1}{0.0127} \approx 78.74\text{ cm}.

Explanation:

When a lens is immersed in a denser medium like water, its refractive power decreases, leading to an increase in its focal length (approximately 44 times for glass in water).