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Physics - Refraction through Lenses and Optical Instruments

Grade 10ICSE

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 curved. A convex lens is a converging lens (thicker at the middle), while a concave lens is a diverging lens (thinner at the middle).

Optical Centre (OO): The point on the principal axis at the centre of the lens. A ray of light passing through OO undergoes no deviation.

Principal Focus (FF): For a convex lens, it is the point where rays parallel to the principal axis actually meet (Real Focus). For a concave lens, it is the point from which they appear to diverge (Virtual Focus).

Focal Length (ff): The distance between the optical centre and the principal focus. By convention, ff is positive for a convex lens and negative for a concave lens.

Rules for Ray Diagrams: (1) A ray parallel to the principal axis passes through (convex) or appears to come from (concave) the focus. (2) A ray passing through the optical centre goes straight. (3) A ray passing through or directed towards the focus emerges parallel to the principal axis.

Nature of Images (Convex Lens): Real and inverted for all positions of object except when the object is between F1F_1 and OO. When u<fu < f, the image is virtual, erect, and magnified.

Nature of Images (Concave Lens): The image formed is always virtual, erect, and diminished, regardless of the position of the object.

Sign Convention (New Cartesian): All distances are measured from the optical centre. Distances in the direction of incident light are positive. Heights above the principal axis are positive.

Magnification (mm): The ratio of the height of the image (hih_i) to the height of the object (hoh_o). It is also equal to the ratio of image distance (vv) to object distance (uu).

Power of a Lens (PP): It is the measure of the deviation (convergence or divergence) produced by a lens. It is the reciprocal of the focal length in metres. The S.I. unit is Dioptre (DD).

📐Formulae

1v1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

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

P=1f (in metres)=100f (in cm)P = \frac{1}{f \text{ (in metres)}} = \frac{100}{f \text{ (in cm)}}

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

💡Examples

Problem 1:

An object is placed at a distance of 20 cm20\text{ cm} from a convex lens of focal length 10 cm10\text{ cm}. Find the position and nature of the image.

Solution:

Given: u=20 cmu = -20\text{ cm} (sign convention), f=+10 cmf = +10\text{ cm} (convex lens). Using the lens formula: 1v120=110    1v+120=110    1v=110120=2120=120\frac{1}{v} - \frac{1}{-20} = \frac{1}{10} \implies \frac{1}{v} + \frac{1}{20} = \frac{1}{10} \implies \frac{1}{v} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20}. Therefore, v=+20 cmv = +20\text{ cm}. Magnification m=vu=2020=1m = \frac{v}{u} = \frac{20}{-20} = -1.

Explanation:

Since vv is positive, the image is formed on the other side of the lens at 20 cm20\text{ cm}. Since m=1m = -1, the image is real, inverted, and the same size as the object (object is at 2F12F_1).

Problem 2:

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

Solution:

Given: f=15 cmf = -15\text{ cm} (concave lens), v=10 cmv = -10\text{ cm} (concave lens always forms a virtual image on the same side as the object). Using the lens formula: 1101u=115    1u=115+110=2+330=130\frac{1}{-10} - \frac{1}{u} = \frac{1}{-15} \implies -\frac{1}{u} = -\frac{1}{15} + \frac{1}{10} = \frac{-2+3}{30} = \frac{1}{30}. Therefore, u=30 cmu = -30\text{ cm}.

Explanation:

The object should be placed at a distance of 30 cm30\text{ cm} in front of the concave lens.

Problem 3:

Calculate the power of a convex lens of focal length 25 cm25\text{ cm}.

Solution:

Given: f=+25 cm=+0.25 mf = +25\text{ cm} = +0.25\text{ m}. Using the formula P=1f(m)P = \frac{1}{f(m)}: P=10.25=+4 DP = \frac{1}{0.25} = +4\text{ D}.

Explanation:

The power is positive because the lens is convex (converging).