Optics Question 482

Question: A 60 watt bulb is hung over the center of a table 4 m x 4 m at a height of 3 m. The ratio of the intensities of illumination at a point on the centre of the edge and on the corner of the table is

[CPMT 1976, 84]

Options:

A) $ {{(17/13)}^{3/2}} $

B) 2 / 1

C) 17 / 13

D) 5 / 4

Show Answer

Answer:

Correct Answer: A

Solution:

The illuminance at A is $ I _{A}=\frac{L}{{{(\sqrt{13})}^{2}}}\times \cos {\theta _{1}}=\frac{L}{13}\times \frac{3}{\sqrt{13}}=\frac{3L}{{{(13)}^{3/2}}} $.

The illuminance at B is $ I _{B}=\frac{L}{{{(\sqrt{17})}^{2}}}\times \cos {\theta _{2}} $

$ =\frac{L}{17}\times \frac{3}{\sqrt{17}}=\frac{3L}{{{(17)}^{3/2}}} $

$ \therefore $ $ \frac{I _{A}}{I _{B}}={{( \frac{17}{13} )}^{3/2}} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक