Solveeit Logo

Question

Question: Derive \( R = 2f \) for a spherical mirror, where the symbols have their usual meaning....

Derive R=2fR = 2f for a spherical mirror, where the symbols have their usual meaning.

Explanation

Solution

Hint : First law of reflection states that the incident ray, reflected ray and normal, all lie in the same plane. Second law of reflection states that the angle of reflection is always equal to the angle of incidence. The angles are measured relative to the perpendicular to the surface at the point where the ray strikes the surface.

Complete Step By Step Answer:
Let OAOA be the incident ray and CACA be the line normal to the surface at the point AA where the ray (OA)\left( {OA} \right) strikes the surface of a spherical mirror. Let ff and RR be the focal length and radius of curvature. Also let ii and rr be the angle of incidence and the angle of reflection. Then we can draw the figure as follows

Hence FP=fFP = f and CP=RCP = R .
Also CC is the centre of the circle and RR is also the radius of the sphere.
According to second law of reflection
i=ri = r ----(1)
Since OAOA is parallel to CPCP we get !ACP=!OAC=i\left| \\!{\underline {\, {ACP} \,}} \right. = \left| \\!{\underline {\, {OAC} \,}} \right. = i ---(2)
From the equation (1), the equation (2) becomes
!ACP=i=r=!CAF\left| \\!{\underline {\, {ACP} \,}} \right. = i = r = \left| \\!{\underline {\, {CAF} \,}} \right.
Then ΔACF\Delta ACF is an isosceles triangle. Therefore CF=FACF = FA ---(3)
For mirror of small aperture we get FAFPFA \approx FP ---(4)
From the equation (3) and (4), we get
CF=FPCF = FP ---(5)
then CP=CF+FP=FP+FP=2  FPCP = CF + FP = FP + FP = 2\;FP
R=2f\Rightarrow R = 2f .
Hence the proof.

Note :
Myopia(near-sightedness) is corrected by using a Concave Lens of suitable power. Hypermetropia(farsightedness) is corrected by using a convex lens of suitable power. A spherical mirror is a mirror which has the shape of a piece cut out of a spherical surface. There are two types of spherical mirrors: concave, and convex.