ray of light

Dear student,

 

Using snell's law,

 

n ref.ind X 0.7 = n air X sin(angle)

 

1.6 X 0.7 = 1 X sin (angle)

 

Sin -1 ( 1.12 ) = 1 X angle

 

This is greater than 90 deg.

 

i here is 45 deg.

 

Hence, The sine of the angle of refraction of the emergent ray will be greater than 0.7.

 

Hope this helps.

Thanking you

Team

Topperelarning.com