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Light – Reflection and Refraction - Reflection of light by curved surfaces

Grade 10CBSE

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

🔑Concepts

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Spherical Mirrors: Reflecting surfaces which are part of a hollow sphere. They are classified into Concave (converging) and Convex (diverging) mirrors.

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Important Terms: Pole (PP) is the center of the reflecting surface; Center of Curvature (CC) is the center of the sphere of which the mirror is a part; Radius of Curvature (RR) is the distance PCPC; Principal Focus (FF) is the point where parallel rays converge or appear to diverge from.

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Relationship between RR and ff: For spherical mirrors of small apertures, the radius of curvature is found to be twice the focal length, expressed as R=2fR = 2f.

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New Cartesian Sign Convention: The object is always placed to the left of the mirror. Distances measured in the direction of incident light are positive, while those measured against it are negative. Height upwards (perpendicular to principal axis) is positive (+h+h), and downwards is negative (−h-h).

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Concave Mirror Image Formation: Depending on the position of the object (uu), it can form real and inverted images (magnified, diminished, or same size) or a virtual and erect image (when the object is between PP and FF).

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Convex Mirror Image Formation: It always forms a virtual, erect, and diminished image behind the mirror, regardless of the object's position.

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Magnification (mm): It is the ratio of the height of the image (h′h') to the height of the object (hh). A negative mm indicates a real image, while a positive mm indicates a virtual image.

📐Formulae

R=2fR = 2f

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

m=h′h=−vum = \frac{h'}{h} = -\frac{v}{u}

💡Examples

Problem 1:

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, nature, and size of the image.

Solution:

Given: Radius of curvature R=+3.00 mR = +3.00 \text{ m}, Object distance u=−5.00 mu = -5.00 \text{ m}. Focal length f=R2=+3.002=+1.50 mf = \frac{R}{2} = \frac{+3.00}{2} = +1.50 \text{ m}. Using mirror formula: 1v+1u=1f  ⟹  1v=1f−1u=11.50−1−5.00=11.50+15.00=5.00+1.507.50=6.507.50\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{1.50} - \frac{1}{-5.00} = \frac{1}{1.50} + \frac{1}{5.00} = \frac{5.00 + 1.50}{7.50} = \frac{6.50}{7.50}. Therefore, v=7.506.50=+1.15 mv = \frac{7.50}{6.50} = +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 a distance of 1.15 m1.15 \text{ m} behind the mirror. Since vv is positive and mm is positive (+0.23+0.23), the image is virtual, erect, and diminished (smaller by a factor of 0.230.23).

Problem 2:

An object 4.0 cm4.0 \text{ cm} in size is placed at 25.0 cm25.0 \text{ cm} in front of a concave mirror of focal length 15.0 cm15.0 \text{ cm}. At what distance from the mirror should a screen be placed in order to obtain a sharp image?

Solution:

Given: Object size h=+4.0 cmh = +4.0 \text{ cm}, Object distance u=−25.0 cmu = -25.0 \text{ cm}, Focal length f=−15.0 cmf = -15.0 \text{ cm} (concave mirror). Using mirror formula: 1v=1f−1u=1−15.0−1−25.0=−115.0+125.0=−5+375=−275\frac{1}{v} = \frac{1}{f} - \frac{1}{u} = \frac{1}{-15.0} - \frac{1}{-25.0} = -\frac{1}{15.0} + \frac{1}{25.0} = \frac{-5 + 3}{75} = \frac{-2}{75}. Thus, v=−37.5 cmv = -37.5 \text{ cm}. Height of image h′=m⋅h=(−vu)h=(−−37.5−25.0)×4.0=−1.5×4.0=−6.0 cmh' = m \cdot h = (-\frac{v}{u})h = (-\frac{-37.5}{-25.0}) \times 4.0 = -1.5 \times 4.0 = -6.0 \text{ cm}.

Explanation:

The screen should be placed at 37.5 cm37.5 \text{ cm} from the mirror on the same side as the object. The image is real (since vv is negative), inverted (since h′h' is negative), and enlarged.

Reflection of light by curved surfaces Revision - Class 10 Science CBSE