A coin lies at the bottom of a 0.87 m -deep pool. Part A If a viewer sees it at

Question

A coin lies at the bottom of a 0.87 m -deep pool.

Part A

If a viewer sees it at a 33 degree angle, where is the image of the coin, relative to the coin? [Hint: The image is found by tracing back to the intersection of two rays.]

Assume that x-axis is directed horizontally toward the viewer, and y-axis is directed vertically upward. Express your answer using two significant figures. Enter the coordinates separated by a comma.

A coin lies at the bottom of a 0.87 m -deep pool.

Part A

If a viewer sees it at a 33 degree angle, where is the image of the coin, relative to the coin? [Hint: The image is found by tracing back to the intersection of two rays.]

Assume that x-axis is directed horizontally toward the viewer, and y-axis is directed vertically upward. Express your answer using two significant figures. Enter the coordinates separated by a comma.

Explanation / Answer

angle of incidence = 33 degrees

refractive index of air and water are 1 and 1.33 respectively

so according to snells law we get

n1sin(theta1) = n2sin(theta2)

1*sin33 = 1.33sin(theta2) we get theta2 as 24.174 degrees.

tan(theta1)/tan(theta2) = Dr/Da= real depth/apparent depth

tan(33)/tan(27.174) = 0.87/x

apparent depth = x = 0.688 m

tan(theta2) = A/Dr

A = 0.446 m

therefore the coordinates of the image are (0.45 m , -0.69 m)

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