A silver medallion is sealed within a transparent block of plastic. An observer in air, viewing the medallion from directly above, sees the medallion at an apparent depth of 2.0 cm beneath the top surface of the block. How far below the top surface would the medallion appear if the observer (not wearing goggles) and the block were under water (n

2.66 cm

Explanation:

Apparent depth ( d1 ) = 2.0 cm

d2 = ?

Determine how far below top surface the medallion would appear

lets express The two scenarios

Case 1 : apparent depth as d1 = dactual (Nair/ nplastic) . ( eyes in the air )

Case 2 : Apparent depth as d2 = dactual (Nwater / nplastic) ( eyes in water )

Final How far below top surface will the medallion appear

d2 = d1 * Nwater / Nair

    = ( 2 * 1.33) / 1  = 2.66 cm

note :  Nwater = 1.33,  , Nair = 1 ( missing values in your question )

the components of a vector are the parts of the vectors that are to each other.

b perpendicular