Problem Statement
A point object is placed 20 cm from the surface of a glass sphere of diameter $D = 10$ cm ($n = 1.5$). Find the final image position.
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Treat as two refractions at spherical surfaces ($R = 5$ cm).
Step 2 — Apply the relevant physical law or equation: First surface (air to glass, $R_1 = +5$ cm, $u_1 = -20$ cm):
Step 3 — Solve algebraically for the unknown: $$\frac{n_2}{v_1}-\frac{n_1}{u_1}=\frac{n_2-n_1}{R_1} \implies \frac{1.5}{v_1}+\frac{1}{20}=\frac{0.5}{5}=0.1$$
$$\frac{1.5}{v_1}=0.05 \implies v_1=30\text{ cm (inside glass, beyond sphere)}$$
Step 4 — Substitute numerical values with units: Second surface (glass to air, $R_2 = -5$ cm). Object for 2nd surface: $u_2 = 30-10 = 20$ cm inside glass, so $u_2 = +20$ cm (on same side as incoming light):
Step 5 — Compute and check the result: $$\frac{1}{v_2}-\frac{1.5}{20}=\frac{1-1.5}{-5}=0.1$$
$$\frac{1}{v_2}=0.1+0.075=0.175 \implies v_2 = 5.7\text{ cm}$$
Step 6: Final image is $\approx 5.7$ cm to the right of the far surface of the sphere.
Worked Calculation
$$\frac{n_2}{v_1}-\frac{n_1}{u_1}=\frac{n_2-n_1}{R_1} \implies \frac{1.5}{v_1}+\frac{1}{20}=\frac{0.5}{5}=0.1$$
$$\frac{1.5}{v_1}=0.05 \implies v_1=30\text{ cm (inside glass, beyond sphere)}$$
$$\frac{1}{v_2}-\frac{1.5}{20}=\frac{1-1.5}{-5}=0.1$$
Treat as two refractions at spherical surfaces ($R = 5$ cm).
First surface (air to glass, $R_1 = +5$ cm, $u_1 = -20$ cm):
$$\frac{n_2}{v_1}-\frac{n_1}{u_1}=\frac{n_2-n_1}{R_1} \implies \frac{1.5}{v_1}+\frac{1}{20}=\frac{0.5}{5}=0.1$$
$$\frac{1.5}{v_1}=0.05 \implies v_1=30\text{ cm (inside glass, beyond sphere)}$$
Second surface (glass to air, $R_2 = -5$ cm). Object for 2nd surface: $u_2 = 30-10 = 20$ cm inside glass, so $u_2 = +20$ cm (on same side as incoming light):
$$\frac{1}{v_2}-\frac{1.5}{20}=\frac{1-1.5}{-5}=0.1$$
$$\frac{1}{v_2}=0.1+0.075=0.175 \implies v_2 = 5.7\text{ cm}$$
Final image is $\approx 5.7$ cm to the right of the far surface of the sphere.
Answer
$$\boxed{\frac{1}{v_2}=0.1+0.075=0.175 \implies v_2 = 5.7\text{ cm}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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