Question:
Figure shows an irregular block of material of refractive index \(\sqrt{2}\). A ray of light strikes the face AB as shown in figure. After refraction, it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at E. Find the distance OE upto two places of decimal.
Figure shows an irregular block of material of refractive index \(\sqrt{2}\). A ray of light strikes the face AB as shown in figure. After refraction, it is incident on a spherical surface CD of radius of curvature 0.4 m and enters a medium of refractive index 1.514 to meet PQ at E. Find the distance OE upto two places of decimal.
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When a ray travels parallel to the axis and hits a spherical refracting surface, the object distance is effectively infinity. Use $\mu_2/v = (\mu_2-\mu_1)/R$.
Updated On: Apr 20, 2026
- 7 m
- 7.29 m
- 6.06 m
- 8.55 m
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
Apply Snell's law at face AB and then the refraction formula at spherical surface CD: \(\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}\).
Step 2: Detailed Explanation:
At AB: \(\sin 60° / \sin r = \sqrt{2}/1 \Rightarrow \sin r = 1/2 \Rightarrow r = 30°\). The refracted ray inside the block becomes parallel to AD (travels horizontally). At spherical surface CD: object at \(u = \infty\), \(\mu_1 = \sqrt{2}\), \(\mu_2 = 1.514\), \(R = 0.4\) m: \[ \frac{1.514}{v} = \frac{1.514 - \sqrt{2}}{0.4} = \frac{1.514 - 1.414}{0.4} = \frac{0.1}{0.4} \] \[ v = \frac{1.514 \times 0.4}{0.1} = 6.056 \approx 6.06 \text{ m} \]
Step 3: Final Answer:
Distance OE \(= 6.06\) m.
Apply Snell's law at face AB and then the refraction formula at spherical surface CD: \(\dfrac{\mu_2}{v} - \dfrac{\mu_1}{u} = \dfrac{\mu_2 - \mu_1}{R}\).
Step 2: Detailed Explanation:
At AB: \(\sin 60° / \sin r = \sqrt{2}/1 \Rightarrow \sin r = 1/2 \Rightarrow r = 30°\). The refracted ray inside the block becomes parallel to AD (travels horizontally). At spherical surface CD: object at \(u = \infty\), \(\mu_1 = \sqrt{2}\), \(\mu_2 = 1.514\), \(R = 0.4\) m: \[ \frac{1.514}{v} = \frac{1.514 - \sqrt{2}}{0.4} = \frac{1.514 - 1.414}{0.4} = \frac{0.1}{0.4} \] \[ v = \frac{1.514 \times 0.4}{0.1} = 6.056 \approx 6.06 \text{ m} \]
Step 3: Final Answer:
Distance OE \(= 6.06\) m.
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