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Light: Reflection and Refraction - REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT

Grade 10CBSE

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

🔑Concepts

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Reflection of light is the phenomenon where light rays strike a polished surface (like a mirror) and bounce back into the same medium.

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The First Law of Reflection states that the angle of incidence ∠i\angle i is always equal to the angle of reflection ∠r\angle r.

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The Second Law of Reflection states that the incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.

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Spherical mirrors are part of a hollow sphere. A Concave mirror has a reflecting surface curved inwards, while a Convex mirror has a reflecting surface curved outwards.

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The relationship between the radius of curvature RR and the focal length ff of a spherical mirror is given by R=2fR = 2f.

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According to the New Cartesian Sign Convention, distances measured in the direction of incident light are positive, while those measured against it are negative. Heights above the principal axis are positive, and below are negative.

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A real image is formed when light rays actually meet after reflection (inverted), while a virtual image is formed when they appear to meet (erect).

📐Formulae

∠i=∠r\angle i = \angle r

f=R2f = \frac{R}{2}

1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

m=hiho=−vum = \frac{h_{i}}{h_{o}} = -\frac{v}{u}

💡Examples

Problem 1:

An object is placed at a distance of 10 cm10\text{ cm} from a convex mirror of focal length 15 cm15\text{ cm}. Find the position and nature of the image.

Solution:

Given: Object distance u=−10 cmu = -10\text{ cm}, Focal length f=+15 cmf = +15\text{ cm} (convex mirror). Using the mirror formula: 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f} 1v+1−10=115\frac{1}{v} + \frac{1}{-10} = \frac{1}{15} 1v=115+110\frac{1}{v} = \frac{1}{15} + \frac{1}{10} 1v=2+330=530\frac{1}{v} = \frac{2 + 3}{30} = \frac{5}{30} v=+6 cmv = +6\text{ cm} Magnification m=−vu=−6−10=+0.6m = -\frac{v}{u} = -\frac{6}{-10} = +0.6.

Explanation:

Since the image distance vv is positive, the image is formed behind the mirror at 6 cm6\text{ cm}. The positive magnification m<1m < 1 indicates the image is virtual, erect, and diminished.

Problem 2:

A concave mirror produces a three times magnified real image of an object placed at 10 cm10\text{ cm} in front of it. Where is the image located?

Solution:

Given: Object distance u=−10 cmu = -10\text{ cm}. Since the image is real and magnified, m=−3m = -3. Using the magnification formula: m=−vum = -\frac{v}{u} −3=−v−10-3 = -\frac{v}{-10} −3=v10-3 = \frac{v}{10} v=−30 cmv = -30\text{ cm}

Explanation:

The image is located at a distance of 30 cm30\text{ cm} in front of the mirror. The negative sign indicates it is a real image.

REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT9.1 REFLECTION OF LIGHT9.1…