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Light – Reflection and Refraction - Lens formula and magnification

Grade 10CBSE

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

🔑Concepts

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The Lens Formula expresses the relationship between the object distance (uu), image distance (vv), and the focal length (ff) of a spherical lens, given by 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}.

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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 and image distance as m=vum = \frac{v}{u}.

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According to the New Cartesian Sign Convention, the focal length of a convex lens is always taken as positive (+f+f), while the focal length of a concave lens is always negative (−f-f).

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The object distance (uu) is always taken as negative because the object is placed to the left of the lens.

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A positive magnification (+m+m) indicates a virtual and erect image, while a negative magnification (−m-m) indicates a real and inverted image.

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Power of a lens (PP) is the reciprocal of its focal length in meters (P=1fP = \frac{1}{f}). The SI unit of power is Dioptre (DD).

📐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)}}

Pnet=P1+P2+P3+...P_{net} = P_1 + P_2 + P_3 + ...

💡Examples

Problem 1:

A convex lens has a focal length of 10 cm10\text{ cm}. At what distance should an object be placed from the lens so that it forms a real and inverted image 20 cm20\text{ cm} away on the other side? Also, find the magnification.

Solution:

Given: f=+10 cmf = +10\text{ cm} (convex lens), v=+20 cmv = +20\text{ cm} (real image is on the opposite side). Using the lens formula: 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}. Substituting values: 120−1u=110  ⟹  1u=120−110  ⟹  1u=1−220=−120\frac{1}{20} - \frac{1}{u} = \frac{1}{10} \implies \frac{1}{u} = \frac{1}{20} - \frac{1}{10} \implies \frac{1}{u} = \frac{1 - 2}{20} = -\frac{1}{20}. Thus, u=−20 cmu = -20\text{ cm}. Magnification m=vu=20−20=−1m = \frac{v}{u} = \frac{20}{-20} = -1.

Explanation:

The object is placed at 2f2f (20 cm20\text{ cm}), resulting in an image of the same size (∣m∣=1|m|=1) but inverted (mm is negative).

Problem 2:

A concave lens has a focal length of 15 cm15\text{ cm}. If it forms an image 10 cm10\text{ cm} from the lens, calculate the object distance and the power of the lens.

Solution:

Given: f=−15 cmf = -15\text{ cm}, v=−10 cmv = -10\text{ cm} (concave lens always forms virtual images on the same side). 1v−1u=1f  ⟹  −110−1u=−115  ⟹  1u=115−110  ⟹  1u=2−330=−130\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \implies -\frac{1}{10} - \frac{1}{u} = -\frac{1}{15} \implies \frac{1}{u} = \frac{1}{15} - \frac{1}{10} \implies \frac{1}{u} = \frac{2 - 3}{30} = -\frac{1}{30}. So, u=−30 cmu = -30\text{ cm}. Power P=1f(m)=1−0.15 m≈−6.67 DP = \frac{1}{f(m)} = \frac{1}{-0.15\text{ m}} \approx -6.67\text{ D}.

Explanation:

The object is placed 30 cm30\text{ cm} in front of the lens. The negative power indicates a diverging (concave) lens.

Lens formula and magnification - Revision Notes & Key Formulas | CBSE Class 10 Science