Complex apparent and real depth problemm

[pistonsfan321]

Homework Statement

A small logo is embedded in a thick block of transparent material (n = 1.73), 3.59 cm beneath the top surface of the block. The block is put under water (n = 1.333), so there is 1.86 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?

Homework Equations

apparent depth = dn_2/(n_1)

The Attempt at a Solution

What i tried doing is applying this equation twice. First with glass and water 3.59(1.33/1.73). Then with water and air 1.86(1/1.33). After that I added the two values of the apparent distances which I got 4.162

 

[Doc Al]

What i tried doing is applying this equation twice. First with glass and water 3.59(1.33/1.73).

OK. That's the apparent location of the first image below the block surface. What's the location of that image below the water surface? That's the distance you want when using the equation the second time.

 

[nlightner]

cm.

This approach is a good start, but it is important to consider the different refractive indices at each interface (glass-water and water-air) separately. The equation you used is for calculating the apparent depth when light travels from a medium with a higher refractive index (n1) to a medium with a lower refractive index (n2). In this case, the light travels from air (n1 = 1) to water (n2 = 1.333) and then from water (n1 = 1.333) to glass (n2 = 1.73). This means you need to use different equations for each interface.

To calculate the apparent depth when light travels from air to water, you can use the equation you mentioned: apparent depth = dn2/n1. Plugging in the values, we get:

apparent depth = (1.86 cm)(1.333)/1 = 2.481 cm

This means that the logo appears to be 2.481 cm below the surface of the water when viewed from above.

To calculate the apparent depth when light travels from water to glass, we can use a similar equation: apparent depth = dn1/n2. Plugging in the values, we get:

apparent depth = (3.59 cm)(1.333)/1.73 = 2.767 cm

This means that the logo appears to be 2.767 cm below the surface of the glass when viewed from above.

To find the total apparent depth, we simply add these two values together:

total apparent depth = 2.481 cm + 2.767 cm = 5.248 cm

Therefore, the logo appears to be 5.248 cm below the surface of the water when viewed from above. It is important to note that this is the apparent depth, which is different from the real depth (the actual distance from the top surface of the water to the logo). The real depth can be calculated using the equation: real depth = apparent depth x n2/n1. In this case, the real depth would be 3.59 cm, which is the same as the original depth of the logo in the glass block. This highlights the complex nature of apparent and real depth in situations involving multiple interfaces and different refractive indices.