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Optics - Refraction of Light at Plane and Spherical Surfaces

Grade 12ICSEPhysics

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

🔑Concepts

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Refraction is the change in direction of light as it passes from one transparent medium to another due to a change in its speed. The absolute refractive index is given by μ=cv\mu = \frac{c}{v}.

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Snell's Law: For a given pair of media and a given color of light, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant, i.e., sin⁡isin⁡r=1μ2\frac{\sin i}{\sin r} = {}^1\mu_2.

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Principle of Reversibility: If the path of a ray of light is reversed, it retraces its original path, expressed as aμb=1bμa{}^a\mu_b = \frac{1}{{}^b\mu_a}.

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Real and Apparent Depth: Due to refraction, an object placed in a denser medium appears closer when viewed from a rarer medium. The refractive index is μ=Real DepthApparent Depth\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}.

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Total Internal Reflection (TIR): When light travels from a denser to a rarer medium at an angle of incidence greater than the critical angle (i>Ci > C), it is reflected back into the denser medium.

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Refraction at Spherical Surfaces: When light travels from a medium of refractive index μ1\mu_1 to μ2\mu_2 through a surface of radius of curvature RR, the relationship between object distance uu and image distance vv is governed by the spherical surface formula.

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Lens Maker's Formula: Relates the focal length ff of a lens to the refractive index μ\mu of the material and the radii of curvature R1R_1 and R2R_2 of its two surfaces.

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Power of a Lens: It is the ability of a lens to converge or diverge light rays, defined as P=1fP = \frac{1}{f} (where ff is in meters). The S.I. unit is Dioptre (DD).

📐Formulae

μ=cv\mu = \frac{c}{v}

n1sin⁡θ1=n2sin⁡θ2n_1 \sin \theta_1 = n_2 \sin \theta_2

μ=Real DepthApparent Depth\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}

sin⁡C=1μ\sin C = \frac{1}{\mu}

μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}

1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

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

m=vu=hihom = \frac{v}{u} = \frac{h_i}{h_o}

P=1f(m)=100f(cm)P = \frac{1}{f(m)} = \frac{100}{f(cm)}

Peq=P1+P2−dP1P2P_{eq} = P_1 + P_2 - d P_1 P_2

💡Examples

Problem 1:

A tank is filled with water to a height of 12.5 cm12.5\text{ cm}. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm9.4\text{ cm}. What is the refractive index of water?

Solution:

Given: Real depth h=12.5 cmh = 12.5\text{ cm}, Apparent depth h′=9.4 cmh' = 9.4\text{ cm}. Using the formula μ=Real DepthApparent Depth\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}, we get μ=12.59.4≈1.33\mu = \frac{12.5}{9.4} \approx 1.33.

Explanation:

Refractive index is the ratio of the actual thickness of the medium to the observed thickness when viewed normally.

Problem 2:

A biconvex lens has radii of curvature 20 cm20\text{ cm} and 30 cm30\text{ cm}. If the refractive index of the glass is 1.51.5, find its focal length.

Solution:

Using Lens Maker's Formula: 1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right). Given μ=1.5\mu = 1.5, R1=+20 cmR_1 = +20\text{ cm}, R2=−30 cmR_2 = -30\text{ cm}. 1f=(1.5−1)(120−1−30)=0.5(120+130)=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{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:

Note the sign convention: R1R_1 is positive for the first surface and R2R_2 is negative for the second surface of a biconvex lens.

Problem 3:

A point object is placed in air at a distance of 40 cm40\text{ cm} from a concave spherical glass surface (μ=1.5\mu = 1.5) of radius of curvature 20 cm20\text{ cm}. Find the position of the image.

Solution:

Using μ2v−μ1u=μ2−μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. Here μ1=1\mu_1 = 1 (air), μ2=1.5\mu_2 = 1.5 (glass), u=−40 cmu = -40\text{ cm}, R=−20 cmR = -20\text{ cm} (concave). 1.5v−1−40=1.5−1−20⇒1.5v+140=−0.520⇒1.5v=−140−140=−240=−120\frac{1.5}{v} - \frac{1}{-40} = \frac{1.5 - 1}{-20} \Rightarrow \frac{1.5}{v} + \frac{1}{40} = -\frac{0.5}{20} \Rightarrow \frac{1.5}{v} = -\frac{1}{40} - \frac{1}{40} = -\frac{2}{40} = -\frac{1}{20}. Thus, v=1.5×(−20)=−30 cmv = 1.5 \times (-20) = -30\text{ cm}.

Explanation:

The negative sign indicates that the image is virtual and formed on the same side as the object.

Refraction of Light at Plane and Spherical Surfaces Revision - Class 12 Physics ICSE