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प्रश्न
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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
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संबंधित प्रश्न
A ray of light is incident on a plane mirror at an angle of 30°. What is the angle of reflection
Name the spherical mirror which has:
(a) virtual principal focus.
(b) real principal focus.
Name the spherical mirror which can produce a virtual and diminished image of an object.
A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image.
When an object of height 1 cm is kept at a distance 4 cm from a concave mirror, its erect image of height 1.5 cm is formed at a distance 6 cm behind the mirror. Find the focal length of mirror, by drawing.
What do you understand by the term real image?
What is the relationship between the focal length and radius of curvature?
Complete the following diagrams shown in the below figure by drawing the reflected ray for each incident ray.

Define the following terms in the context of a diverging mirror:
- Principal focus
- Focal length
Draw a labelled ray diagram to illustrate your answer.
