(A) A triangular glass prism of thickness 12cm is placed on a mark on a piece of paper resting on a horizontal bench. If the refractive index of the material of the prism is 1.5, calculate the apparent displacement of the mark.

(B)Water is poured into a jar to a depth of 24cm. The bottom of the jar appears
to be raised by 6cm when viewed vertically. Calculate the refractive index of water.

Since the refractive index of the triangular glass prism is 1.5, n = 1.5. Also, the triangular glass prism is 12 cm thick and placed on a mark on a piece of paper resting on a horizontal bench. So, the real depth of the mark on the piece of paper through the triangular prism is D = 12 cm and its apparent depth is L.

Since n = D/L,

making L, subject of the formula, we have

L = D/n

Substituting the values of the variables into the equation, we have

L = D/n

L = 12 cm/1.5

L = 8 cm

so, the apparent depth of the mark is 8 cm.

So, the apparent displacement of the mark is d = D - L = 12 cm - 8 cm = 4 cm

Thus the apparent displacement of the mark is 4 cm.

Part B - The refractive index of water, n = 1.33.

The refractive index of a material, n = real depth, D/apparent depth, L.

So, n = D/L

Since the depth of water in the jar is 24 cm, the real depth, D = 24 cm.

Also, the bottom of the jar appears to be raised by 6 cm. So, the apparent depth, L = 24 cm - 6 cm = 18 cm

Since n = D/L, the refractive index of water n = D/L

Substituting the values of the variables into the equation, we have

n = D/L

= 24 cm/18 cm

= 1.33

So, the refractive index of water, n = 1.33.

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