Relationship Between Focal Length and (NEET Physics)

Write the relation between focal length and radius of curvature of a spherical mirror.

Hint: Before stating the relation, derive the relation between radius of curvature and the focal length of a mirror. This will help you understand better. Consider that when a beam of parallel light strikes a concave mirror, it converges at the mirror’s focal point after reflection.

Complete step by step answer:

The relationship between the focal length f and the radius of curvature R of a spherical mirror states that the focal length is half the radius of curvature, expressed as: [latex]  ƒ = \frac{R}{2}[/latex] Let us derive this relation.

Imagine a concave mirror where the radius of curvature significantly exceeds the diameter of its aperture.

We can take help from the figure given.

Suppose a parallel ray of light is incident on the mirror as shown in the figure. Let this incident ray make an angle θ with normal to the surface of the mirror. As this mirror is a segment of a circle, the normal line drawn to its surface intersects at the center (C) of the mirror. It is known that rays of light parallel to the principal axis converge at the focus (F) of a concave mirror after reflection. According to the reflection principle, the angle of incidence is equal to the angle of reflection.

Therefore, ∠BAC=∠FAC=θ as shown in the given figure.

Since BA and PC are parallel, ∠BAC=∠ACF=θ.

Therefore, from exterior angle theorem ∠AFN=∠ACF+∠CAF=2θ.

Now drop a normal CP from point A. Let the foot of this normal be N.

Here, tanθ=[latex]\frac{AN}{CN}[/latex] ⇒ AN = CN tanθ ……. (i).

tan 2θ = [latex]\frac{AN}{FN}[/latex] ⇒ AN = FN tan 2θ ….. (ii).

From equation (i) and equation (ii) we get, CN tanθ = FN tan 2θ

⇒ [latex]\frac{CN}{FN}[/latex]= [latex]\frac{tan2θ}{tanθ}[/latex] …… (iii).

Since the radius of curvature is very much larger than the diameter of its aperture, NP is very small compared to CN and CP and θ will be a small angle .

Therefore, CN ≈ CP and FN ≈ FP.

For small angles tan⁡θ = θ and tan⁡2 θ = 2θ.

Therefore, equation (iii) can be written as

⇒ [latex]\frac{CP}{FP}[/latex]=[latex]\frac{2θ}{θ}[/latex] ⇒ FP = [latex]\frac{CP}{2}[/latex]

And  CP =  R  and  FP = f.

Hence,  ƒ  =[latex]\frac{R}{2}[/latex]

Note: Note that this relation between the radius of curvature (R) of a concave mirror and the focal length (ƒ) of the mirror, which is [latex]ƒ = \frac{R}{2}[/latex], is true only when the R is very much larger than the diameter of its aperture.