Question

an object is placed at the distance of 10 centimetre from convex mirror of focal length 15cm find the position and nature of the image​

Answer 1

Answer:

-30distance of the image

Explanation:

1/f=1/u+1/v [mirror formula ]

=1/15-1/10=1/v

=-1/30=1/v

=-30

Answer 2

given, u=-10cm,f=-15cm

now ,using mirror formula,

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ = > - \frac{1}{15} = \frac{1}{v} +( - \frac{1}{10} ) \\ = > - \frac{1}{15} = \frac{1}{v} - \frac{1}{10} \\ = > \frac{1}{v} = - \frac{1}{15} + \frac{1}{10} \\ = > \frac{1}{v} = \frac{ - 2 + 3}{30} \\ = > \frac{1}{v} = \frac{ 1}{30} = > v = 30$

thus,image will be formed at a distance of 30cm in behind the mirror. the image is virtual and erect .

hope this helps.

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