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Light – Reflection and Refraction - Mirror formula and magnification

Grade 10CBSE

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

🔑Concepts

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New Cartesian Sign Convention: The pole (PP) of the mirror is taken as the origin. The object is always placed to the left of the mirror, making the object distance (uu) always negative.

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Focal Length (ff): According to sign convention, the focal length of a concave mirror is always negative, while the focal length of a convex mirror is always positive.

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Mirror Formula: It defines the relationship between the object distance (uu), image distance (vv), and the focal length (ff) of a spherical mirror.

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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 expressed as the negative ratio of image distance to object distance (m=−v/um = -v/u).

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Nature of Image from mm: If mm is negative, the image is real and inverted. If mm is positive, the image is virtual and erect.

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Size of Image from mm: If ∣m∣>1|m| > 1, the image is enlarged. If ∣m∣<1|m| < 1, the image is diminished. If ∣m∣=1|m| = 1, the image is the same size as the object.

📐Formulae

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

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

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

💡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: R=+3.00 mR = +3.00\text{ m}, u=−5.00 mu = -5.00\text{ m}.

  1. Focal length f=R2=+3.002=+1.50 mf = \frac{R}{2} = \frac{+3.00}{2} = +1.50\text{ m}.
  2. 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}.
  3. 1v=5.00+1.507.50=6.507.50  ⟹  v=7.506.50=+1.15 m\frac{1}{v} = \frac{5.00 + 1.50}{7.50} = \frac{6.50}{7.50} \implies v = \frac{7.50}{6.50} = +1.15\text{ m}.
  4. 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 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: h=4.0 cmh = 4.0\text{ cm}, u=−25.0 cmu = -25.0\text{ cm}, f=−15.0 cmf = -15.0\text{ cm}.

  1. Using Mirror Formula: 1v=1f−1u=1−15.0−1−25.0=−115.0+125.0\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}.
  2. 1v=−5+375=−275  ⟹  v=−37.5 cm\frac{1}{v} = \frac{-5 + 3}{75} = \frac{-2}{75} \implies v = -37.5\text{ cm}.
  3. Magnification m=−vu=−−37.5−25.0=−1.5m = -\frac{v}{u} = -\frac{-37.5}{-25.0} = -1.5.
  4. 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. The negative sign of vv indicates a real image. The magnification m=−1.5m = -1.5 and height h′=−6.0 cmh' = -6.0\text{ cm} show the image is real, inverted, and enlarged.