Q. A thin glass plate of refractive index 1.5 is placed over a sheet of paper on which there is a small dot. The thickness of the plate is 3 cm. When viewed from directly above, the dot appears to be displaced by:

अपवर्तनांक 1.5 की एक पतली कांच की प्लेट को कागज की एक शीट पर रखा गया है जिस पर एक छोटा बिंदु है। प्लेट की मोटाई 3 सेमी है। जब सीधे ऊपर से देखा जाता है, तो बिंदु विस्थापित होता हुआ प्रतीत होता है:

A
1 cm
B
2 cm
C
3 cm
D
1.5 cm

Explanation

The apparent depth dad_a when viewed through a medium with refractive index nn is given by:

da=dnd_a = \frac{d}{n}

where d=3 cmd = 3 \text{ cm} is the real thickness of the plate, and n=1.5

da=31.5=2 cmd_a = \frac{3}{1.5} = 2 \text{ cm}

The displacement Δd=d−da=3−2=1 cm\Delta d = d - d_a = 3 - 2 = 1 \text{ cm}.
But the question asks how much the dot appears displaced, so the displacement is 1 cm, not 2 cm.
Correction: The displacement = d−da=3−2=1 cmd - d_a = 3 - 2 = 1 \text{ cm}
So the dot appears to be displaced by 1 cm.

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