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Light: Reflection and Refraction - SPHERIC9.2 SPHERIC9.2 SPHERIC9.2 SPHERIC9.2 SPHERIC AL MIRRORSAL MIRRORSAL MIRRORSAL MIRRORSAL MIRRORS

Grade 10CBSE

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

🔑Concepts

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A spherical mirror is a mirror whose reflecting surface is a part of a hollow sphere of glass. They are categorized into two types: Concave Mirror (reflecting surface curved inwards) and Convex Mirror (reflecting surface curved outwards).

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The Center of Curvature (CC) is the center of the sphere of which the mirror is a part. The Radius of Curvature (RR) is the distance between the Pole (PP) and the Center of Curvature (CC).

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The Principal Focus (FF) is a point on the principal axis where rays parallel to the principal axis meet (concave) or appear to diverge from (convex) after reflection. The distance PFPF is the Focal Length (ff).

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For spherical mirrors of small apertures, the radius of curvature is found to be 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 are positive, while those opposite are negative. Distances above the principal axis are positive, and below are negative.

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For a Concave Mirror, focal length ff is always negative. For a Convex Mirror, focal length ff is always positive.

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Mirror Formula: It provides the relationship between the object distance (uu), image distance (vv), and focal length (ff) as 1f=1v+1u\frac{1}{f} = \frac{1}{v} + \frac{1}{u}.

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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). It is also related to uu and vv as m=−vum = -\frac{v}{u}. If mm is negative, the image is real; if mm is positive, the image is virtual.

📐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:

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: Height of object 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 the mirror formula: 1v=1f−1u\frac{1}{v} = \frac{1}{f} - \frac{1}{u} 1v=1−15.0−1−25.0\frac{1}{v} = \frac{1}{-15.0} - \frac{1}{-25.0} 1v=−115+125=−5+375=−275\frac{1}{v} = -\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} in front of the mirror. Since vv is negative, the image is real. Since mm is negative and height is −6.0 cm-6.0\text{ cm}, the image is inverted and 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}, 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=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=5.00+1.507.50=6.507.50\frac{1}{v} = \frac{1}{1.50} + \frac{1}{5.00} = \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}

Explanation:

The image is formed at a distance of 1.15 m1.15\text{ m} at the back of the mirror. Since vv is positive, the image is virtual and erect. Because it is a convex mirror, the image is also diminished.