Question 118: Ray Optics

Question

A plano-convex lens having radius of curvature of first surface 2 cm exhibits focal length of $f_1$ in air. Another plano-convex lens with first surface radius of curvature 3 cm has focal length of $f_2$ when it is immersed in a liquid of refractive index 1.2. If both the lenses are made of same glass of refractive index 1.5 , the ratio of $f_1$ and $f_2$ will be

Accepted Answer

$$ \frac{1}{f} = \left(\frac{n}{n'} - 1ight)\left(\frac{1}{R_1} - \frac{1}{R_2}ight) $$
For a thin plano-convex lens, one surface is flat, meaning $$R_2 = \infty$$ and $$\frac{1}{R_2} = 0.$$ The formula reduces to:
$$ \frac{1}{f} = \left(\frac{n}{n'} - 1ight)\frac{1}{R_1} $$
Thus, the focal length is:
$$ f = \frac{R_1}{\left(\frac{n}{n'} - 1ight)} $$
For the First Lens (in Air)
Refractive index of lens, $$n = 1.5$$
Surrounding medium (air), $$n' = 1$$
Radius of curvature, $$R_1 = 2 	ext{ cm}$$
Substitute into the formula:
$$ f_1 = \frac{2}{\left(\frac{1.5}{1} - 1ight)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 	ext{ cm} $$
For the Second Lens (in Liquid)
Refractive index of lens, $$n = 1.5$$
Surrounding medium (liquid), $$n' = 1.2$$
Radius of curvature, $$R_1 = 3 	ext{ cm}$$
Substitute into the formula:
$$ f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1ight)} $$
First, calculate the term inside the parentheses:
$$ \frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25 $$
Now, compute $$f_2$$:
$$ f_2 = \frac{3}{0.25} = 12 	ext{ cm} $$
Ratio of the Focal Lengths
$$ \frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3} $$
Thus, the ratio of $$f_1$$ to $$f_2$$ is:
$$ 1 : 3 $$

Answer

$2: 3$

Answer

$1: 3$

Answer

$1: 2$

Answer

$3: 5$

Ray Optics

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