Homework Help

A layer of benzene (n=1.50) 2.60 cm deep floats on water (n=1.33) that is 6.50cm deep....

user profile pic

durbanville | High School Teacher | (Level 1) Educator Emeritus

Posted August 30, 2013 at 12:55 PM via web

dislike 1 like

A layer of benzene (n=1.50) 2.60 cm deep floats on water (n=1.33) that is 6.50cm deep. What is the apparent distance from the upper benzene surface to the bottom of the water layer when it is viewed at normal incidence? Please explain.

Thank you. 

1 Answer | Add Yours

Top Answer

user profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted August 30, 2013 at 7:10 PM (Answer #1)

dislike 2 like

Consider refraction through the water layer (layer 2) into the benzene layer (layer 3) first (Please see the attached diagram). Further assume air to be the layer 1.

n(2,3)= n(2,1)/n(3,1)=1.33/1.50=0.886667

OT is the real depth, after refraction through the water layer.
PT is the apparent depth after refraction through the water layer into the benzene layer.
We know that,
Refractive index n = Real depth/Apparent depth
Thus,
`n = (OT)/(PT)`
`rArr 0.886667 = 6.5/(PT)`

`PT = 6.5/0.886667=7.33` cm

Now consider refraction through the benzene layer into the air.

PW is the real depth, after refraction through the benzene layer.
QW is the apparent depth after refraction through the benzene layer into the air.
Plugging in the values in this case,

`1.5 = (PW)/(QW)`
`rArr QW = (PW)/1.5`

`=(PT+TW)/1.5` cm

`=(7.33+2.6)/1.5 ` cm

`= 6.62 ` cm (when viewed from normal incidence).

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes