Section 26.2 A small logo is embedded in a thick block of crown glass (n 1.52),

Question

Section 26.2 A small logo is embedded in a thick block of crown glass (n 1.52), 3.10 cm beneath the top surface of the glass. The block is put under water, so there is 1.20 cm of water above the top surface of the block. The logo is viewed from directly above by an observer in air. How far beneath the top surface of the water does the logo appear to be? 242 X cm Section 26.2 Answers CJ10 26.P 028. A point source of light is submerged 2.6 m below the surface of a lake and emits rays in all directions. On the surface of the lake, directly above the source, the area illuminated is a circle. What is the maximum radius that this circle could have? Additional Materials section 26.3

Explanation / Answer

To an observer in the water directly above the logo, the apparent depth of the logo is

dg1 = dg*(nw/ng)

When viewed directly from above in air, the logo’s apparent depth is,

dw1 = (dw + dg1)*(nair/nw)

using both equations:

dw1 = [dw + dg*(nw/ng)]*(nair/nw)

dw1 = dw*(nair/nw) + dg*(nair/ng)

dw1 = 1.20*(1/1.333) + 3.1*(1/1.52)

dw1 = 2.94 cm

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