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Ray Optics and Optical Instruments - Reflection of Light by Spherical Mirrors

Grade 12CBSEPhysics

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

🔑Concepts

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Laws of Reflection: (i) The angle of incidence ∠i\angle i equals the angle of reflection ∠r\angle r. (ii) The incident ray, the reflected ray, and the normal at the point of incidence all lie in the same plane.

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New Cartesian Sign Convention: The pole PP is taken as the origin. Distances measured in the direction of incident light are positive (++), while those opposite to it are negative (−-).

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Focal Length and Radius of Curvature: For spherical mirrors with small apertures, the focal length ff is half the radius of curvature RR, expressed as f=R2f = \frac{R}{2}.

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Concave Mirror: A converging mirror where the focal length ff is always taken as negative. It can form both real and virtual images depending on the object position.

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Convex Mirror: A diverging mirror where the focal length ff is always taken as positive. It always forms a virtual, erect, and diminished image regardless of the object's position.

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Linear Magnification mm: It is the ratio of the height of the image hih_i to the height of the object hoh_o. mm is negative for real images and positive for virtual images.

📐Formulae

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

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

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

m=ff−um = \frac{f}{f - u}

v=ufu−fv = \frac{uf}{u - f}

💡Examples

Problem 1:

An object of height 5.0 cm5.0\text{ cm} is placed at a distance of 20.0 cm20.0\text{ cm} in front of a concave mirror of radius of curvature 30.0 cm30.0\text{ cm}. Find the position, nature, and size of the image.

Solution:

Given: height of object ho=5.0 cmh_o = 5.0\text{ cm}, object distance u=−20.0 cmu = -20.0\text{ cm}, radius of curvature R=−30.0 cmR = -30.0\text{ cm}. First, calculate focal length: f=R2=−30.02=−15.0 cmf = \frac{R}{2} = \frac{-30.0}{2} = -15.0\text{ cm} Using the mirror formula 1v+1u=1f\frac{1}{v} + \frac{1}{u} = \frac{1}{f}: 1v=1f−1u=1−15−1−20\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15} - \frac{1}{-20} 1v=−115+120=−4+360=−160\frac{1}{v} = -\frac{1}{15} + \frac{1}{20} = \frac{-4 + 3}{60} = -\frac{1}{60} So, v=−60.0 cmv = -60.0\text{ cm}. Magnification m=−vu=−−60−20=−3m = -\frac{v}{u} = -\frac{-60}{-20} = -3. Height of image hi=m×ho=−3×5=−15.0 cmh_i = m \times h_o = -3 \times 5 = -15.0\text{ cm}. Calculation for the difference in height magnitude: 15−510\begin{array}{r} 15 \\ - 5 \\ \hline 10 \end{array}

Explanation:

The negative sign of vv indicates the image is formed 60.0 cm60.0\text{ cm} in front of the mirror (Real image). The negative sign of hih_i indicates the image is inverted. Since ∣m∣>1|m| > 1, the image is magnified. The image is three times larger than the object, showing a height difference of 10 cm10\text{ cm} in magnitude.

Problem 2:

A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m3.00\text{ m}. If a bus is located at 5.00 m5.00\text{ m} from this mirror, find the position and magnification of the image.

Solution:

Given: R=+3.00 mR = +3.00\text{ m} (convex), u=−5.00 mu = -5.00\text{ m}. Focal length f=R2=+1.50 mf = \frac{R}{2} = +1.50\text{ m}. Using mirror formula: 1v=1f−1u=11.50−1−5.00=11.5+15\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1.50} - \frac{1}{-5.00} = \frac{1}{1.5} + \frac{1}{5} 1v=1015+315=1315\frac{1}{v} = \frac{10}{15} + \frac{3}{15} = \frac{13}{15} v=1513≈1.15 mv = \frac{15}{13} \approx 1.15\text{ m} Magnification m=−vu=−1.15−5.00=+0.23m = -\frac{v}{u} = -\frac{1.15}{-5.00} = +0.23.

Explanation:

The image is formed at 1.15 m1.15\text{ m} behind the mirror. The positive magnification confirms the image is virtual and erect. The value 0.230.23 indicates the image is diminished to 23%23\% of the original size.

Reflection of Light by Spherical Mirrors Class 12 Notes & Examples