A coin lies on the bottom of a pool under 1.5 m of water and 0.90 m from the sid

Question

A coin lies on the bottom of a pool

under 1.5 m of water and 0.90 m from the sidewall (Fig.22.28).

(a) If a light beam is incident on the water surface atthe wall, at

what angle relative tothe wall must the beam be directed

so that it will illuminate the coin?

(b) What is the apparent depth d’ of the coin?Do not use the

expression for the apparent depth derived with thesmall

angle approximation.

(c) Calculate the apparent depth using d’ = d/n(derived with

small angle). Calculate a percent difference withyour

Explanation / Answer

tan1=0.9/1.5 1=30.963756 Applying Snells law we have as n1sini=n2sinr 1*sin=1.33*sin30.963756 =43.178957974 Apparent depth d1=1.5/1.33=1.1278m Hence we get by it.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---