Question:
An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is N/100, where N is an integer, the value of N is ________.
An observer can see through a small hole on the side of a jar (radius 15 cm) at a point at height of 15 cm from the bottom (see figure). The hole is at a height of 45 cm. When the jar is filled with a liquid up to a height of 30 cm the same observer can see the edge at the bottom of the jar. If the refractive index of the liquid is N/100, where N is an integer, the value of N is ________.
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Jar diameter is 30 cm. The empty-jar ray makes \(45^\circ\) with the vertical; after filling, the ray in the liquid reaches the bottom edge with \(\tan r = 15/30\).
Updated On: Nov 5, 2026
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Correct Answer: 158
Solution and Explanation
The jar diameter is 30 cm. Without liquid the line of sight goes from the hole (height 45 cm, one wall) to the opposite wall at height 15 cm: horizontal 30 cm, vertical drop 30 cm, so it makes \(45^\circ\) with the vertical (normal to the liquid surface). Thus \(\sin i = 1/\sqrt2\).
With liquid, the same ray in air reaches the liquid surface (height 30 cm, i.e. 15 cm below the hole) after moving 15 cm horizontally. In the liquid it must reach the bottom edge of the opposite wall: it descends 30 cm and moves the remaining \(30-15 = 15\) cm horizontally, so \(\tan r = 15/30 = 1/2\) and \(\sin r = 1/\sqrt5\).
Snell's law: \(n = \sin i/\sin r = \sqrt5/\sqrt2 = \sqrt{2.5} \approx 1.58\).
\(N = 100n \approx 158\).
Answer: 158
With liquid, the same ray in air reaches the liquid surface (height 30 cm, i.e. 15 cm below the hole) after moving 15 cm horizontally. In the liquid it must reach the bottom edge of the opposite wall: it descends 30 cm and moves the remaining \(30-15 = 15\) cm horizontally, so \(\tan r = 15/30 = 1/2\) and \(\sin r = 1/\sqrt5\).
Snell's law: \(n = \sin i/\sin r = \sqrt5/\sqrt2 = \sqrt{2.5} \approx 1.58\).
\(N = 100n \approx 158\).
Answer: 158
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