
Two mirrors, each of 1.6 m long are facing each other. The distance between the mirror is 20 cm. A light ray incident on one end of one of the mirrors at an angle of incidence of $30°$. How many times is the ray reflected before it reaches the other end?
Answer
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Hint: Firstly, use the law of reflection to find the angle of reflection of the light from the angle of incidence. Then, find the horizontal distance covered by the light ray for one reflection. Now, to find the total number of reflections, take the ratio of length of the mirror and horizontal distance covered by the light ray for one reflection. This obtained number of times the ray is reflected before reaching the other end.
Complete answer:
Given: BD= 20 cm
DY= BX= 1.6 m= 160 cm
According to angle of reflection,
$\angle ABD= \angle CBD= 30°$
In $\triangle BDC$,
$\tan { 30°}=\dfrac { DC }{ BD } $
Substituting values in above equation we get,
$\dfrac { 1 }{ \sqrt { 3 } } =\dfrac { DC }{ 20 }$
$\Rightarrow DC=\dfrac { 20 }{ \sqrt { 3 } }$
$\Rightarrow DC= 11.55 cm$
Thus, the ray of light undergoes one reflection for every horizontal distance of 11.55 cm.
The total distance covered to make all the reflection will be given by,
$N=\dfrac { Length\quad of\quad mirror }{ Distance\quad covered\quad for\quad one\quad reflection }$
$\Rightarrow N= \dfrac {DY}{DC}$
Substituting values in above equation we get,
$N= \dfrac {160}{11.55}$
$\Rightarrow N= 13.85$
$\Rightarrow N \approx 14$
Thus, the light ray is reflected 14 times before it reaches the other end.
Note:
Students must make sure that when they are substituting the values in an equation, all the involved physical quantities are expressed in their respective S.I. units. If we increase the angle of incidence of light ray then the horizontal distance covered by the light ray to undergo one reflection will be more. Thus, the light will reflect a smaller number of times before reaching the other end of the mirror.
Complete answer:
Given: BD= 20 cm
DY= BX= 1.6 m= 160 cm
According to angle of reflection,
$\angle ABD= \angle CBD= 30°$
In $\triangle BDC$,
$\tan { 30°}=\dfrac { DC }{ BD } $
Substituting values in above equation we get,
$\dfrac { 1 }{ \sqrt { 3 } } =\dfrac { DC }{ 20 }$
$\Rightarrow DC=\dfrac { 20 }{ \sqrt { 3 } }$
$\Rightarrow DC= 11.55 cm$
Thus, the ray of light undergoes one reflection for every horizontal distance of 11.55 cm.
The total distance covered to make all the reflection will be given by,
$N=\dfrac { Length\quad of\quad mirror }{ Distance\quad covered\quad for\quad one\quad reflection }$
$\Rightarrow N= \dfrac {DY}{DC}$
Substituting values in above equation we get,
$N= \dfrac {160}{11.55}$
$\Rightarrow N= 13.85$
$\Rightarrow N \approx 14$
Thus, the light ray is reflected 14 times before it reaches the other end.
Note:
Students must make sure that when they are substituting the values in an equation, all the involved physical quantities are expressed in their respective S.I. units. If we increase the angle of incidence of light ray then the horizontal distance covered by the light ray to undergo one reflection will be more. Thus, the light will reflect a smaller number of times before reaching the other end of the mirror.
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