Question

When a concave mirror is placed facing the sun, the sun's rays converge to a point 10cm from the mirror. Now, an erect, 2-cm-long pin is placed 15cm away on the principal axis of the mirror. If you want to get the image of the pin on a card, where would you place the card? What would be the nature and height of the image?

Answer 1

obj should be placed at centre of curvature so the img would be formed at centre of curvature . the img size would be same size of the obj and it is real and inverted
guess this helps you

Answer 2

Answer:

v = - 20

$\display \text{h _i = -2 cm}$

Explanation:

Given :

Focal length = 10 cm

Object distance = 20 cm

We have image formula :

1 / f = 1 / v + 1 / u

1 / v = 1 / - 10 + 1 / 20

1 / v = - 1  20

v = - 20.

Therefore , screen would be placed at distance of 20 cm to get image.

We have :

m = - v / u = h_i / h_o

- 20 / 20 = h_i / 2

h_i = - 2 cm.

Hence size of the image is 2 cm.