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Light – Reflection and Refraction - Refraction of light by spherical lens

Grade 10CBSE

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

🔑Concepts

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A lens is a transparent material bound by two surfaces, of which one or both surfaces are spherical. A convex lens is thicker at the middle and is called a converging lens, while a concave lens is thicker at the edges and is called a diverging lens.

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The central point of a lens is its optical center (OO). Lines passing through the centers of curvature of the two surfaces of the lens is the principal axis.

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For a convex lens, rays parallel to the principal axis converge at the principal focus (FF). For a concave lens, they appear to diverge from the principal focus (FF).

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The distance between the optical center and the principal focus is the focal length (ff).

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Sign Convention: Distances measured in the direction of incident light are positive. Object distance (uu) is always negative. Focal length (ff) is positive for a convex lens and negative for a concave lens.

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Magnification (mm) produced by a lens is the ratio of the height of the image (h′h') to the height of the object (hh). It is also related to the object distance (uu) and image distance (vv).

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Power of a lens (PP) is the reciprocal of its focal length in meters. The SI unit of power is Dioptre (DD). 1D=1m−11 D = 1 m^{-1}.

📐Formulae

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

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

💡Examples

Problem 1:

A concave lens has a focal length of 15 cm15 \text{ cm}. At what distance should the object from the lens be placed so that it forms an image at 10 cm10 \text{ cm} from the lens? Also, find the magnification produced by the lens.

Solution:

Given: f=−15 cmf = -15 \text{ cm} (concave lens), v=−10 cmv = -10 \text{ cm} (concave lens forms a virtual image on the same side). Using the lens formula: 1v−1u=1f  ⟹  1u=1v−1f  ⟹  1u=1−10−1−15=−110+115=−3+230=−130\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies \frac{1}{u} = \frac{1}{v} - \frac{1}{f} \implies \frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} = -\frac{1}{10} + \frac{1}{15} = \frac{-3 + 2}{30} = -\frac{1}{30}. Thus, u=−30 cmu = -30 \text{ cm}. Magnification m=vu=−10−30=+13≈0.33m = \frac{v}{u} = \frac{-10}{-30} = +\frac{1}{3} \approx 0.33.

Explanation:

The negative sign of uu indicates the object is placed 30 cm30 \text{ cm} in front of the lens. The positive magnification less than 11 shows the image is virtual, erect, and diminished.

Problem 2:

Find the power of a convex lens of focal length 20 cm20 \text{ cm}.

Solution:

Given: f=+20 cm=+0.2 mf = +20 \text{ cm} = +0.2 \text{ m}. Using the power formula: P=1f=10.2=+5 DP = \frac{1}{f} = \frac{1}{0.2} = +5 \text{ D}.

Explanation:

Power is the reciprocal of focal length in meters. A convex lens has a positive focal length, resulting in a positive power of +5+5 Dioptres.

Refraction of light by spherical lens Revision - Class 10 Science CBSE