Question

an object is placed at a distance of 20 cm from a convex mirror of focal length 15 cm. find the position and nature of the image

Answer 1

$\bf\huge u = -20cm , f = 15cm$

$\bf\huge\frac{1}{f} = \frac{1}{u}+\frac{1}{v}$

$\bf\huge\frac{1}{v} = \frac{1}{15} - \frac{1}{-20}$

$\bf\huge\frac{1}{v} = \frac{1}{15} - \frac{1}{20}$

$\bf\huge\frac{1}{v} = \frac{7}{60}$

$\bf\huge\texttt v = 8.57 cm$

$\bf\huge\boxed{\boxed{\:Nature\:of\:Image\:=\:Virtual}}}$

Answer 2

Explanation:

v

1

+

u

1

=

f

1

⇒

v

1

+

−20

1

=

10

−1

⇒

v

1

=−

10

1

+

20

1

=

20

−2+1

=

20

−1

v=−20cm

The object is placed at center of curvature & hence will him image at

′

c

′

.