Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर

< i = < r
But < ABC = alternate< BCF
Therefore< CBF = < BCF
And the ΔFBC is isosceles
BF = FC ... (i)
PF = FC
PF + PF= PF + FC 2PF = PC
Now since PF = f, the focal length of the mirror
And PC = R, the radius of curvature of the mirror
From here we can determine the focal length of the concave mirror i.e. half of radius of curvature
APPEARS IN
संबंधित प्रश्न
One half of a convex lens of focal length 10 cm is covered with a black paper. Can such a lens produce an image of a complete object placed at a distance of 30 cm from the lens? Draw a ray diagram to justify your answer. A 4 cm tall object is placed perpendicular to its principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 15 cm. Find the nature, position and the size of the image.
Define the following term in the context of spherical mirrors:- Centre of curvature
What happens when a ray of light falls normally (or perpendiculary) on the surface of a plane mirror?
When a spherical mirror is held towards the sun and its sharp image is formed on a piece of a carbon paper for some time, a hole is burnt in the carbon paper.
What name is given to the distance between spherical mirror and carbon paper?
In each case (a) and (b), draw reflected rays for the given incident rays and mark focus by the symbol F.

A concave mirror forms a virtual image of size twice that of the object placed at a distance 5 cm from it.
Find : (a) the focal length of the mirror (b) position of image
An object is placed at 4 cm distance in front of a concave mirror of radius of curvature 24 cm. Find the position of image. Is the image magnified?
Define the following term in relation to concave mirror.
Principal focus
The image formed in a plane mirror is always inverted.
Define the following terms in the context of a diverging mirror:
- Principal focus
- Focal length
Draw a labelled ray diagram to illustrate your answer.
