Question
A container is filled with water $$\left( {\mu = 1.33} \right)$$ upto a height of $$33.25\,cm.$$ A concave mirror is placed $$15\,cm$$ above the water level and the image of an object placed at the bottom is formed $$25\,cm$$ below the water level. Focal length of the mirror is
A container is filled with water $$\left( {\mu = 1.33} \right)$$ upto a height of $$33.25\,cm.$$ A concave mirror is placed $$15\,cm$$ above the water level and the image of an object placed at the bottom is formed $$25\,cm$$ below the water level. Focal length of the mirror is
A.
$$15\,cm$$
B.
$$20\,cm$$
C.
$$-18.31\,cm$$
D.
$$10\,cm$$
Answer :
$$-18.31\,cm$$
Solution :
The image $$I'$$ for first refraction (i.e., when the ray comes out of liquid) is at a depth of
$$ = \frac{{33.25}}{{1.33}} = 25\,cm\,\,\left[ {\because {\text{Apparent depth}} = \frac{{{\text{Real depth}}}}{\mu }} \right]$$
Now, reflection will occur at concave mirror. For this $$I'$$ behaves as an object.
$$\therefore u = - \left( {15 + 25} \right) = - 40\,cm\,\,{\text{and}}\,\,v = - \left[ {15 + \frac{{25}}{{1.33}}} \right]$$
Where $$\frac{{25}}{{1.33}}$$ is the real depth of the image.
Using mirror formula we get
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} + \frac{1}{u}, \cr & f = - 18.31\,cm \cr} $$
The image $$I'$$ for first refraction (i.e., when the ray comes out of liquid) is at a depth of
$$ = \frac{{33.25}}{{1.33}} = 25\,cm\,\,\left[ {\because {\text{Apparent depth}} = \frac{{{\text{Real depth}}}}{\mu }} \right]$$
Now, reflection will occur at concave mirror. For this $$I'$$ behaves as an object.
$$\therefore u = - \left( {15 + 25} \right) = - 40\,cm\,\,{\text{and}}\,\,v = - \left[ {15 + \frac{{25}}{{1.33}}} \right]$$
Where $$\frac{{25}}{{1.33}}$$ is the real depth of the image.
Using mirror formula we get
$$\eqalign{ & \frac{1}{f} = \frac{1}{v} + \frac{1}{u}, \cr & f = - 18.31\,cm \cr} $$
