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Ray Optics and Optical Instruments - Refraction and Total Internal Reflection

Grade 12CBSEPhysics

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

🔑Concepts

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Refraction is the phenomenon of change in the path of light as it goes from one medium to another. The basic cause is the change in the speed of light: v=cnv = \frac{c}{n}.

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Snell's Law: For a given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, expressed as n1sin⁡i=n2sin⁡rn_1 \sin i = n_2 \sin r.

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Refractive Index (nn): It is the ratio of the speed of light in a vacuum (c≈3×108 m/sc \approx 3 \times 10^8 \text{ m/s}) to the speed of light in the medium (vv).

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Lateral Shift: The perpendicular distance between the incident ray and the emergent ray when light passes through a rectangular glass slab.

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Apparent Depth: When an object is in a denser medium and viewed from a rarer medium, it appears to be at a shallower depth dapp=drealnd_{app} = \frac{d_{real}}{n}.

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Total Internal Reflection (TIR): When light travels from an optically denser medium to a rarer medium and the angle of incidence is greater than the critical angle (i>ici > i_c), the ray is reflected back into the denser medium.

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Conditions for TIR: 1. Light must travel from a denser to a rarer medium. 2. The angle of incidence must exceed the critical angle ici_c.

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Critical Angle (ici_c): The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90∘90^\circ. It is given by sin⁡ic=1n\sin i_c = \frac{1}{n}.

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Applications of TIR: Optical fibers (used in telecommunications), mirage formation in deserts, and the brilliance of diamonds.

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Lens Maker's Formula: Relates the focal length (ff) of a lens to the refractive index (nn) and the radii of curvature (R1,R2R_1, R_2) of its surfaces.

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Power of a Lens (PP): The ability of a lens to converge or diverge light rays, measured as the reciprocal of focal length in meters (P=1fP = \frac{1}{f}).

📐Formulae

n21=n2n1=v1v2=sin⁡isin⁡rn_{21} = \frac{n_2}{n_1} = \frac{v_1}{v_2} = \frac{\sin i}{\sin r}

n=cvn = \frac{c}{v}

Normal Shift=t(1−1n)\text{Normal Shift} = t \left( 1 - \frac{1}{n} \right)

sin⁡ic=n2n1 (where n1>n2)\sin i_c = \frac{n_2}{n_1} \text{ (where } n_1 > n_2 \text{)}

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)

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})}

Ptotal=P1+P2+P3+…P_{total} = P_1 + P_2 + P_3 + \dots

💡Examples

Problem 1:

A ray of light enters a glass slab (ng=1.5n_g = 1.5) from air. If the angle of incidence is 60∘60^\circ, calculate the angle of refraction.

Solution:

Using Snell's law: nasin⁡i=ngsin⁡rn_a \sin i = n_g \sin r. Given na=1n_a = 1, i=60∘i = 60^\circ, and ng=1.5n_g = 1.5. 1⋅sin⁡60∘=1.5⋅sin⁡r1 \cdot \sin 60^\circ = 1.5 \cdot \sin r 32=1.5sin⁡r  ⟹  sin⁡r=33=1.7323≈0.577\frac{\sqrt{3}}{2} = 1.5 \sin r \implies \sin r = \frac{\sqrt{3}}{3} = \frac{1.732}{3} \approx 0.577 r=sin⁡−1(0.577)≈35.2∘r = \sin^{-1}(0.577) \approx 35.2^\circ

Explanation:

Light bends towards the normal as it moves from a rarer (air) to a denser (glass) medium.

Problem 2:

Calculate the critical angle for a glass-air interface if the refractive index of glass is 1.51.5.

Solution:

The formula for the critical angle is sin⁡ic=1n\sin i_c = \frac{1}{n}. sin⁡ic=11.5=23≈0.6667\sin i_c = \frac{1}{1.5} = \frac{2}{3} \approx 0.6667 ic=sin⁡−1(0.6667)≈41.8∘i_c = \sin^{-1}(0.6667) \approx 41.8^\circ

Explanation:

If light hits the glass-air boundary at an angle greater than 41.8∘41.8^\circ, it will undergo Total Internal Reflection.

Problem 3:

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. Find its focal length.

Solution:

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). By sign convention, R1=+20 cmR_1 = +20\text{ cm} and R2=−30 cmR_2 = -30\text{ cm}. 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)=0.5(3+260)=0.5(560)\frac{1}{f} = 0.5 \left( \frac{1}{20} + \frac{1}{30} \right) = 0.5 \left( \frac{3+2}{60} \right) = 0.5 \left( \frac{5}{60} \right) 1f=12⋅112=124\frac{1}{f} = \frac{1}{2} \cdot \frac{1}{12} = \frac{1}{24} f=24 cmf = 24\text{ cm}

Explanation:

The focal length is positive, indicating that it is a converging lens.

Refraction and Total Internal Reflection Class 12 Notes & Examples