Question

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

Answer 1

given,.

u= -10cm

f= 15cm

by using formula,

1/f= 1/v - 1/u

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

1/15 = 1/v +1/10

1/15- 1/10 = 1/v

v = -30 cm

image form is virtual and image form in front of the mirror and erect.

Answer 2

Answer:

Given -

• Focal length of convex mirror, f = +15 cm
• Object distance, u = -10 cm

According to the mirror formula,

$\\ \implies \sf \: \frac{1}{v} - \frac{1}{f} = \frac{1}{u} \\ \\ \\ \implies \sf \: \frac{1}{v} = \frac{1}{15} - \frac{1}{( - 10)} \\ \\ \\ \implies \sf \: \frac{1}{v} = \frac{1}{15} + \frac{1}{10} \\ \\ \\ \implies \sf \: \frac{1}{v} = \frac{2 + 3}{30} \\ \\ \\ \implies \sf \: \frac{1}{v} = \cancel\frac{5}{30} = 6 \\ \\ \\ \implies \sf \blue{v = 6 \: \: cm} \\ \\ \\ \large{ \underline{ \mathfrak{ \pink{ \: \: \: Magnification \: \: : }}}} \\ \\ \implies \sf \: \frac{ - v}{u} = \frac{ - 6}{ - 10} = 0.6$

The positive and value less than 1 of magnification indicates that the image formed is virtual, errect and diminished.