A one-rupee coin is placed at the bottom of a vessel. Water is then poured into the vessel such that the depth of water becomes 20 cm. If water has refractive index 4/3, the coin would be seen at a depth of

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Q: 37 (NDA-II/2011)
A one-rupee coin is placed at the bottom of a vessel. Water is then poured into the vessel such that the depth of water becomes 20 cm. If water has refractive index 4/3, the coin would be seen at a depth of

question_subject: 

Maths

question_exam: 

NDA-II

stats: 

0,2,3,1,2,2,0

keywords: 

{'rupee coin': [0, 0, 1, 2], 'coin': [0, 3, 6, 4], 'depth': [0, 0, 0, 1], 'refractive index': [0, 1, 1, 5], 'vessel': [2, 0, 3, 3], 'bottom': [2, 2, 2, 7]}

The correct answer is option 3, which is 15 cm. When light travels from one medium (air) to another medium (water), it bends or refracts due to the change in the speed of light in different mediums. This bending of light causes objects to appear shifted or displaced when viewed from outside the water.

The depth at which the coin would be seen can be calculated using the formula:

apparent depth = real depth / refractive index

In this case, the real depth of the coin is 20 cm, and the refractive index of water is 4/3. Plugging these values into the formula, we get:

apparent depth = 20 cm / (4/3) = 20 cm * 3/4 = 15 cm.

Therefore, the coin would be seen at a depth of 15 cm when water with a refractive index of 4/3 is poured into the vessel.