Question

An object is placed at a distance of 10 cm from the pole of a convex mirror of focal lenght 15 cm. Find the nature and position of the image.​

Answer 1

Answer:

Given that,

Object distance, u = –10 cm (By using Cartesian sign convention)

Focal length of a convex mirror, f = + 15 cm (By sign convention)

1/f = 1/u + 1/v

$\\ = > \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ = > \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ = > \frac{1}{v} = \frac{1}{15} - \frac{1}{ - 10} \\ = > v = \frac{150}{25} = 6 \: cm$

Positive sign indicates that the image is virtual.

Magnification = –v/u = –6/–10 = 0.6

Since ‘m’is positive and less than one, the image is erect and diminished.

An erect, virtual, and diminished image is formed at 6 cm from the pole, behind the mirror.

Answer 2

Answer:

Given that,

Object distance, u = –10 cm (By using Cartesian sign convention)

Focal length of a convex mirror, f = + 15 cm (By sign convention)

1/f = 1/u + 1/v

$\\ = > \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ = > \frac{1}{v} = \frac{1}{f} - \frac{1}{u} \\ = > \frac{1}{v} = \frac{1}{15} - \frac{1}{ - 10} \\ = > v = \frac{150}{25} = 6 \: cm$

Positive sign indicates that the image is virtual.

Magnification = –v/u = –6/–10 = 0.6

Since ‘m’is positive and less than one, the image is erect and diminished.

An erect, virtual, and diminished image is formed at 6 cm from the pole, behind the mirror.

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