Question

A ray parallel to the principal axis is incident at ${30^ \circ }$ from a normal concave mirror having radius of curvature $R$ . The point on principal axis where rays are focused is $Q$ such that $PQ$ is

A. $\dfrac{R}{2}$
B. $\dfrac{R}{{\sqrt 3 }}$
C. $\dfrac{{2\sqrt R - R}}{{\sqrt 2 }}$
D. $R\left( {1 - \dfrac{1}{{\sqrt 3 }}} \right)$

Answer 1

Hint: A ray parallel to the main axis will reflect and then pass through the main focus in a concave mirror. With a convex mirror, it appears to veer away from the main focus. After reflection, a ray going through a concave mirror's primary focus will come out parallel to the main axis. The supplied ray will therefore pass through the main focus after the first reflection, and after the second reflection, it will emerge parallel to the main axis.



Formula used:
1. Mirror Formula:
$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}$ … (1)
Here, $f$ is the focal length of the mirror,
$u$ is the object distance and
$v$ is the image distance.
2. Focal length
$f = \dfrac{R}{2}$
where , R is Radius of curvature

Complete answer:
We have been given the data that,
A ray parallel to principal axis is incident at \[{30^ \circ }\] from normal on concave mirror having radius of curvature \[R\]
From the given diagram, we can write as,
From similar triangles, we can write the equation as
\[ = \dfrac{QC}{\sin {{30}^\circ }} = \dfrac{R}{\sin {{120}^\circ }}\]

Now, the equation (1) can be written in terms of \[{\rm{QC}}\] as,
\[{\rm{QC}} = {\rm{R}} \times \dfrac{{\sin {{30}^\circ }}}{{\sin {{120}^\circ }}}\]
We have to solve the above using the identity \[\sin \left( x \right) = \cos \left( {{{90}^{^\circ \:}} - x} \right)\]:
\[ = \dfrac{{\sin \left( {{{30}^{^\circ \:}}} \right)}}{{\cos \left( {{{90}^{^\circ \:}} - {{120}^{^\circ \:}}} \right)}}\]
Let’s use the property \[\cos \left( { - x} \right) = \cos \left( x \right)\]:
\[ = \dfrac{{\sin \left( {{{30}^{^\circ \:}}} \right)}}{{\cos \left( { - {{30}^{^\circ \:}}} \right)}}\]
On simplifying the above, we get
\[ = \dfrac{{\rm{R}}}{{\sqrt 3 }}\]
Thus, we have
\[{\rm{PQ}} = {\rm{PC}} - {\rm{QC}}\]
We can write as,
\[ = {\rm{R}} - \dfrac{{\rm{R}}}{{\sqrt 3 }}\]
We have to take \[{\rm{R}}\] as common, we obtain
\[ = {\rm{R}}\left( {1 - \dfrac{1}{{\sqrt 3 }}} \right)\]
Therefore, the point on principal axis where rays are focused is \[Q\] such that \[PQ\] is \[{\rm{R}}\left( {{\rm{1}} - \dfrac{1}{{\sqrt 3 }}} \right)\]
Hence, the option D is correct.





Note:Students make mistakes in these types of issues, because these types of problems include trigonometry problems. So, they should be careful in applying formulas to get the desired solution.