krit.club logo

Optics - Lenses and Lens Maker's Formula

Grade 12ICSEPhysics

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

🔑Concepts

•

A lens is a transparent refracting medium bounded by two surfaces, at least one of which is a curved surface (usually spherical).

•

Convex Lenses (converging) are thicker at the center than at the edges and have a positive focal length (f>0f > 0).

•

Concave Lenses (diverging) are thinner at the center than at the edges and have a negative focal length (f<0f < 0).

•

The New Cartesian Sign Convention states that the optical center is the origin, distances measured in the direction of incident light are positive, and distances measured against it are negative.

•

The Lens Maker's Formula is used by manufacturers to design lenses of a specific focal length using materials of a known refractive index μ\mu and specific radii of curvature R1R_1 and R2R_2.

•

Linear Magnification (mm) is defined as the ratio of the height of the image (hih_i) to the height of the object (hoh_o). If mm is negative, the image is real and inverted; if mm is positive, the image is virtual and erect.

•

The Power of a lens (PP) is the reciprocal of its focal length in meters, measured in Dioptres (DD). For a convex lens, PP is positive; for a concave lens, PP is negative.

📐Formulae

1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) (Lens Maker's Formula)

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} (Thin Lens Formula)

m=hiho=vum = \frac{h_i}{h_o} = \frac{v}{u} (Magnification)

P=1f (in meters)=100f (in cm)P = \frac{1}{f \text{ (in meters)}} = \frac{100}{f \text{ (in cm)}} (Power of a Lens)

1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} (Effective focal length of two thin lenses in contact)

Ptotal=P1+P2P_{total} = P_1 + P_2 (Total power of lenses in contact)

💡Examples

Problem 1:

An equiconvex lens of refractive index 1.51.5 has a focal length of 20 cm20 \text{ cm} in air. Calculate the radius of curvature of its surfaces.

Solution:

Given: f=+20 cmf = +20 \text{ cm} (for convex), μ=1.5\mu = 1.5. For an equiconvex lens, R1=+RR_1 = +R and R2=−RR_2 = -R. Using the Lens Maker's Formula: 1f=(μ−1)(1R1−1R2)\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) 120=(1.5−1)(1R−1−R)\frac{1}{20} = (1.5 - 1) \left( \frac{1}{R} - \frac{1}{-R} \right) 120=0.5×2R\frac{1}{20} = 0.5 \times \frac{2}{R} 120=1R  ⟹  R=20 cm\frac{1}{20} = \frac{1}{R} \implies R = 20 \text{ cm}

Explanation:

In an equiconvex lens, the two radii of curvature are equal in magnitude but opposite in sign according to the sign convention. Since μ−1=0.5\mu - 1 = 0.5, the focal length ff equals the radius RR for this specific refractive index.

Problem 2:

A convex lens of focal length 10 cm10 \text{ cm} is placed in contact with a concave lens of focal length 20 cm20 \text{ cm}. Find the power and the nature of the combination.

Solution:

Given: f1=+10 cmf_1 = +10 \text{ cm} (convex), f2=−20 cmf_2 = -20 \text{ cm} (concave). The effective focal length FF is: 1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} 1F=110+1−20=2−120=120\frac{1}{F} = \frac{1}{10} + \frac{1}{-20} = \frac{2 - 1}{20} = \frac{1}{20} So, F=+20 cm=0.2 mF = +20 \text{ cm} = 0.2 \text{ m}. Power P=1F (m)=10.2=+5 DP = \frac{1}{F \text{ (m)}} = \frac{1}{0.2} = +5 \text{ D}.

Explanation:

Since the effective focal length FF is positive, the combination behaves like a converging (convex) lens with a total power of +5 Dioptres+5 \text{ Dioptres}.