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Saitama
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Homework Statement
A concave mirror of focal length 30 cm is placed on the flat horizontal surface with its concave side up. Water with refractive index 1.33 is poured into the lens. Where should an object be placed if its image is to be captured on a screen with a magnification of 2.
Ans: 67.5 cm
Homework Equations
The Attempt at a Solution
I have attached a sketch of the given problem. I assume that the downward direction is positive.
The water poured acts as a plano-convex lens. The lens with the concave mirror acts a mirror and the focal length (##f_e##) of this combination is given by:
$$\frac{1}{f_e}=\frac{1}{f_m}-\frac{2}{f_L}$$
where ##f_m## is the focal length of mirror and ##f_L## is focal length of lens.
Calculating focal length of lens:
$$\frac{1}{f_L}=\left(\frac{4}{3}-1\right)\left(\frac{1}{\infty}-\frac{1}{R}\right)$$
where R is the radius of the convex part of lens and I have also approximated 1.33 as 4/3.
Also ##R=2f_m##, hence
$$\frac{1}{f_L}=-\frac{1}{6f_m}$$
Therefore,
$$\frac{1}{f_e}=\frac{4}{3f_m}$$
Since ##f_m=-30\, cm##, ##f_e=-90/4 \, cm##.
The given combination works like a concave mirror. As the image is to be captured on a screen, the image must be real and hence the magnification (m) is -2. For a mirror,
$$m=\frac{-v}{u} \Rightarrow -2=\frac{-v}{u} \Rightarrow v=2u$$
where v is the image distance and u is the object distance from the combination.
$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f_e} \Rightarrow \frac{3}{2u}=\frac{-4}{90}$$
Solving for u gives u=-135/4 cm which is wrong.
Any help is appreciated. Thanks!
