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Light – Reflection and Refraction - Images formed by spherical mirrors

Grade 10CBSE

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

🔑Concepts

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Spherical mirrors are of two types: Concave mirror (reflecting surface curved inwards) and Convex mirror (reflecting surface curved outwards).

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The center of the reflecting surface of a spherical mirror is a point called the Pole (PP). The reflecting surface forms a part of a sphere, the center of which is the Center of Curvature (CC).

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The distance from the Pole to the Principal Focus (FF) is the focal length (ff). For mirrors of small aperture, the radius of curvature (RR) is twice the focal length: 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 (right of the pole) are positive, while those measured against it (left of the pole) are negative. Heights above the principal axis are positive, and below are negative.

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Concave mirrors can produce real and inverted images (for objects beyond FF) or virtual and erect images (when the object is between PP and FF).

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Convex mirrors always produce virtual, erect, and diminished images, regardless of the object's position.

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The linear magnification (mm) is the ratio of the height of the image (h′h') to the height of the object (hh). If mm is negative, the image is real; if mm is positive, the image is virtual.

📐Formulae

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

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

R=2fR = 2f

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

💡Examples

Problem 1:

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? Find the nature and size of the 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}. Using Mirror Formula: 1v=1f−1u\frac{1}{v} = \frac{1}{f} - \frac{1}{u} 1v=1−15.0−1−25.0=−115+125=−5+375=−275\frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0} = -\frac{1}{15} + \frac{1}{25} = \frac{-5 + 3}{75} = \frac{-2}{75} v=−37.5 cmv = -37.5\text{ cm}. Magnification m=−vu=−−37.5−25.0=−1.5m = -\frac{v}{u} = -\frac{-37.5}{-25.0} = -1.5. Height of image h′=m×h=−1.5×4.0=−6.0 cmh' = m \times h = -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. The negative sign of vv indicates the image is real. The negative sign of h′h' indicates the image is inverted. The size 6.0 cm6.0\text{ cm} shows the image is enlarged.

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 nature of the image.

Solution:

Given: R=+3.00 mR = +3.00\text{ m} (convex mirror), so f=R2=+1.50 mf = \frac{R}{2} = +1.50\text{ m}. Object distance u=−5.00 mu = -5.00\text{ m}. Using Mirror Formula: 1v=1f−1u=11.50−1−5.00=11.50+15.00\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} 1v=5.00+1.507.50=6.507.50\frac{1}{v} = \frac{5.00 + 1.50}{7.50} = \frac{6.50}{7.50} 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 but less than 11, the image is virtual, erect, and diminished.