A ray parallel to principal axis is incident at 30° from normal on concave mirror having radius of curvature R. The point on principal axis where rays are focussed is Q such that PQ is

Question

A ray parallel to principal axis is incident at 30° from normal on concave mirror having radius of curvature R. The point on principal axis where rays are focussed is Q such that PQ is (A)  (R/2) (B) (R/√3) (C) (2√R-R/√2) (D) R(1-(1/√2))

Tardigrade Admin · Accepted Answer

From similar triangles, (Q C/ sin 30°)=(R/ sin 120°) or Q C=R × ( sin 30°/ sin 120°)=(R/√3) Thus P Q=P C-Q C=R-(R/√3) =R(1-(1/√3))

Q. A ray parallel to principal axis is incident at $3 0^{\circ}$ from normal on concave mirror having radius of curvature $R$ . The point on principal axis where rays are focussed is $Q$ such that $PQ$ is

$\frac{R}{2}$

$\frac{R}{3}$

$\frac{2 R - R}{2}$

$R (1 - \frac{1}{2})$

From similar triangles,
$\frac{QC}{s i n 3 0 ^{\circ}} = \frac{R}{s i n 12 0 ^{\circ}}$
or $QC = R \times \frac{s i n 3 0 ^{\circ}}{s i n 12 0 ^{\circ}} = \frac{R}{3}$
Thus $PQ = PC - QC = R - \frac{R}{3}$
$= R (1 - \frac{1}{3})$

Q. A ray parallel to principal axis is incident at 30∘ from normal on concave mirror having radius of curvature R. The point on principal axis where rays are focussed is Q such that PQ is

2R​

3​R​

2​2R​−R​

R(1−2​1​)