Answer
Verified
333.1k+ views
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. Use the fact that when a parallel ray of light falls on a concave mirror, it passes through the focus of the mirror after reflection.
Complete step by step answer:
The relation between focal length (f) and radius of curvature (R) of a spherical mirror is that the focal length is equal to half of the radius of curvature i.e. $f=\dfrac{R}{2}$.
Let us derive this relation.
Consider a concave mirror such that its radius of curvature is very much larger than 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 $\theta $ with normal to the surface of the mirror. Since this mirror is part of a circle, the normal drawn to the surface of the circle passes through the centre (C) of the mirror.
We know that rays of light parallel to the principal axis passing through the focus (F) of a concave mirror, after reflection.
According to the law of reflections, the angle of incidence and angle of reflection are equal. Therefore, $\angle BAC=\angle FAC=\theta $ as shown in the given figure.
Since BA and PC are parallel, $\angle BAC=\angle ACF=\theta $.
Therefore, from exterior angle theorem $\angle AFN=\angle ACF+\angle CAF=2\theta $.
Now drop a normal CP from point A. Let the foot of this normal be N.
Here, $\tan \theta =\dfrac{AN}{CN}\Rightarrow AN=CN\tan \theta $ ……. (i).
$\tan 2\theta =\dfrac{AN}{FN}\Rightarrow AN=FN\tan 2\theta $ ….. (ii).
From equation (i) and equation (ii) we get,
$CN\tan \theta =FN\tan 2\theta $
$\Rightarrow \dfrac{CN}{FN}=\dfrac{\tan 2\theta }{\tan \theta }$ …… (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 $\theta $ will be a small angle .
Therefore, $CN\approx CP$ and $FN\approx FP$.
For small angles $\tan \theta =\theta $ and $\tan 2\theta =2\theta $.
Therefore, equation (iii) can be written as
$\Rightarrow \dfrac{CP}{FP}=\dfrac{2\theta }{\theta }\Rightarrow FP=\dfrac{CP}{2}$
And CP=R and FP=f.
Hence, $f=\dfrac{R}{2}$
Note: Note that this relation between the radius of curvature (R) of a concave mirror and the focal length (f) of the mirror, which is $f=\dfrac{R}{2}$, is true only when the R is very much larger than the diameter of its aperture.
Complete step by step answer:
The relation between focal length (f) and radius of curvature (R) of a spherical mirror is that the focal length is equal to half of the radius of curvature i.e. $f=\dfrac{R}{2}$.
Let us derive this relation.
Consider a concave mirror such that its radius of curvature is very much larger than 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 $\theta $ with normal to the surface of the mirror. Since this mirror is part of a circle, the normal drawn to the surface of the circle passes through the centre (C) of the mirror.
We know that rays of light parallel to the principal axis passing through the focus (F) of a concave mirror, after reflection.
According to the law of reflections, the angle of incidence and angle of reflection are equal. Therefore, $\angle BAC=\angle FAC=\theta $ as shown in the given figure.
Since BA and PC are parallel, $\angle BAC=\angle ACF=\theta $.
Therefore, from exterior angle theorem $\angle AFN=\angle ACF+\angle CAF=2\theta $.
Now drop a normal CP from point A. Let the foot of this normal be N.
Here, $\tan \theta =\dfrac{AN}{CN}\Rightarrow AN=CN\tan \theta $ ……. (i).
$\tan 2\theta =\dfrac{AN}{FN}\Rightarrow AN=FN\tan 2\theta $ ….. (ii).
From equation (i) and equation (ii) we get,
$CN\tan \theta =FN\tan 2\theta $
$\Rightarrow \dfrac{CN}{FN}=\dfrac{\tan 2\theta }{\tan \theta }$ …… (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 $\theta $ will be a small angle .
Therefore, $CN\approx CP$ and $FN\approx FP$.
For small angles $\tan \theta =\theta $ and $\tan 2\theta =2\theta $.
Therefore, equation (iii) can be written as
$\Rightarrow \dfrac{CP}{FP}=\dfrac{2\theta }{\theta }\Rightarrow FP=\dfrac{CP}{2}$
And CP=R and FP=f.
Hence, $f=\dfrac{R}{2}$
Note: Note that this relation between the radius of curvature (R) of a concave mirror and the focal length (f) of the mirror, which is $f=\dfrac{R}{2}$, is true only when the R is very much larger than the diameter of its aperture.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Fill the blanks with proper collective nouns 1 A of class 10 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers