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Ray Optics and Optical Instruments - Refraction at Spherical Surfaces and by Lenses

Grade 12CBSEPhysics

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

🔑Concepts

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Refraction at a spherical surface occurs when light travels from a medium of refractive index n1n_1 to another of refractive index n2n_2. The relation between object distance uu, image distance vv, and radius of curvature RR is given by the spherical surface formula.

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The New Cartesian Sign Convention is used: All distances are measured from the pole (PP); distances in the direction of incident light are positive, while those opposite are negative.

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A lens is a transparent medium bound by two surfaces, at least one of which is spherical. A thin lens is one whose thickness is negligible compared to its radii of curvature.

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

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The Thin Lens Formula provides the relationship between the object distance uu, image distance vv, and focal length ff.

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Linear Magnification mm is the ratio of the height of the image h′h' to the height of the object hh. For a lens, m=vum = \frac{v}{u}.

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The Power of a lens PP is the reciprocal of its focal length ff (in meters). It measures the degree of convergence or divergence a lens introduces. The SI unit is Dioptre (DD).

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When two or more thin lenses are placed in contact, the total power PP of the combination is the algebraic sum of the individual powers.

📐Formulae

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

1f=(n21−1)(1R1−1R2)\frac{1}{f} = (n_{21} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) where n21=n2n1n_{21} = \frac{n_2}{n_1}

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

m=h′h=vum = \frac{h'}{h} = \frac{v}{u}

P=1f (in meters)=100f (in cm)P = \frac{1}{f \text{ (in meters)}} = \frac{100}{f \text{ (in cm)}}

1feq=1f1+1f2+…\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} + \dots

Peq=P1+P2+…P_{eq} = P_1 + P_2 + \dots

meq=m1×m2×…m_{eq} = m_1 \times m_2 \times \dots

💡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.

Solution:

Given: R1=+20 cmR_1 = +20\text{ cm}, R2=−30 cmR_2 = -30\text{ cm} (by sign convention), 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) 1f=(1.5−1)(120−1−30)\frac{1}{f} = (1.5 - 1) \left( \frac{1}{20} - \frac{1}{-30} \right) 1f=0.5(120+130)\frac{1}{f} = 0.5 \left( \frac{1}{20} + \frac{1}{30} \right) 1f=0.5(3+260)=0.5×560\frac{1}{f} = 0.5 \left( \frac{3 + 2}{60} \right) = 0.5 \times \frac{5}{60} 1f=2.560=124\frac{1}{f} = \frac{2.5}{60} = \frac{1}{24} Hence, f=24 cmf = 24\text{ cm}.

Explanation:

The Lens Maker's formula is applied here. Note the sign convention where R1R_1 is positive for the first surface of a convex lens and R2R_2 is negative for the second surface.

Problem 2:

Two thin lenses of power +3.5 D+3.5\text{ D} and −1.5 D-1.5\text{ D} are placed in contact. Find the power and focal length of the combination.

Solution:

Given: P1=+3.5 DP_1 = +3.5\text{ D}, P2=−1.5 DP_2 = -1.5\text{ D}. The total power is: P=P1+P2P = P_1 + P_2 P=3.5+(−1.5)=2.0 DP = 3.5 + (-1.5) = 2.0\text{ D} The focal length ff is: f=1P=12.0=0.5 m=50 cmf = \frac{1}{P} = \frac{1}{2.0} = 0.5\text{ m} = 50\text{ cm}

Explanation:

Powers are added algebraically. A positive resulting power indicates the combination behaves as a converging (convex) lens.

Problem 3:

Calculate the image distance for an object placed 10 cm10\text{ cm} in front of a concave lens of focal length 20 cm20\text{ cm}.

Solution:

Given: u=−10 cmu = -10\text{ cm}, f=−20 cmf = -20\text{ cm} (concave lens). Using Lens Formula: 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} 1v=1f+1u\frac{1}{v} = \frac{1}{f} + \frac{1}{u} 1v=1−20+1−10\frac{1}{v} = \frac{1}{-20} + \frac{1}{-10} 1v=−1−220=−320\frac{1}{v} = \frac{-1 - 2}{20} = \frac{-3}{20} v=−203≈−6.67 cmv = -\frac{20}{3} \approx -6.67\text{ cm}

Explanation:

The negative sign of vv indicates that the image is virtual and formed on the same side as the object, which is characteristic of a concave lens.