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Light: Mirrors and Lenses - What Are the Characteristics of Images Formed by Spherical Mirrors?

Grade 8CBSE

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

🔑Concepts

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Spherical mirrors are mirrors whose reflecting surface is a part of a hollow sphere. There are two types: Concave mirrors (reflecting surface curved inwards) and Convex mirrors (reflecting surface curved outwards).

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The geometric center of the reflecting surface is the Pole (PP), and the center of the sphere is the Center of Curvature (CC). The distance PCPC is the Radius of Curvature (RR).

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The Principal Focus (FF) is the point on the principal axis where light rays parallel to the axis converge (concave) or appear to diverge from (convex) after reflection.

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The distance between the Pole and the Principal Focus is the Focal Length (ff). The relationship between RR and ff is given by R=2fR = 2f.

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For a Concave Mirror: The image can be real and inverted (when object is beyond FF) or virtual and erect (when object is between PP and FF). The size varies from highly diminished to highly enlarged.

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For a Convex Mirror: The image is always virtual, erect, and diminished, regardless of the position of the object.

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The Sign Convention: All distances are measured from the Pole. Distances measured in the direction of incident light are positive, while those opposite to incident light are negative. Heights above the principal axis are positive, and below are negative.

📐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}

💡Examples

Problem 1:

An object is placed at a distance of 20 cm20\text{ cm} in front of a concave mirror of focal length 15 cm15\text{ cm}. At what distance from the mirror should a screen be placed to obtain a sharp image?

Solution:

Given: Object distance u=−20 cmu = -20\text{ cm}, Focal length f=−15 cmf = -15\text{ cm}. Using the mirror formula: 1v=1f−1u\frac{1}{v} = \frac{1}{f} - \frac{1}{u} 1v=1−15−1−20\frac{1}{v} = \frac{1}{-15} - \frac{1}{-20} 1v=−115+120\frac{1}{v} = -\frac{1}{15} + \frac{1}{20} 1v=−4+360\frac{1}{v} = \frac{-4 + 3}{60} 1v=−160\frac{1}{v} = -\frac{1}{60} v=−60 cmv = -60\text{ cm}

Explanation:

The screen should be placed at a distance of 60 cm60\text{ cm} on the same side as the object. The negative sign indicates that the image is real and inverted.

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 of the image.

Solution:

Given: R=+3.00 mR = +3.00\text{ m} (convex), so f=R2=+1.50 mf = \frac{R}{2} = +1.50\text{ m}. Object distance u=−5.00 mu = -5.00\text{ m}. 1v=1f−1u\frac{1}{v} = \frac{1}{f} - \frac{1}{u} 1v=11.50−1−5.00\frac{1}{v} = \frac{1}{1.50} - \frac{1}{-5.00} 1v=11.50+15.00\frac{1}{v} = \frac{1}{1.50} + \frac{1}{5.00} 1v=5.00+1.507.50\frac{1}{v} = \frac{5.00 + 1.50}{7.50} 1v=6.507.50\frac{1}{v} = \frac{6.50}{7.50} v=7.506.50≈+1.15 mv = \frac{7.50}{6.50} \approx +1.15\text{ m}

Explanation:

The image is formed at a distance of 1.15 m1.15\text{ m} behind the mirror. Since vv is positive, the image is virtual and erect.

What Are the Characteristics of Images Formed by Spherical Mirrors? Class 8 Notes & Examples