Answer to Question #155804 in Optics for Annmarie

Question #155804

A pin placed in contact with one surface of a glass block of thickness 15ck and of refractive index 1.5 is viewed by the normal of the opposite surface . By how much does the pin appear to be displaced .


1
Expert's answer
2021-01-15T09:32:47-0500

The refractive index can be found with the following equation:


"n=\\frac{d_\\text{real}}{d_\\text{apparent}},\\\\\\space\\\\\nd_\\text{apparent}=\\frac{d_\\text{real}}{n}=\\frac{15}{1.5}=10\\text{ cm}."

So, the pin appear to be displaced by


"d=d_\\text{real}-d_\\text{apparent}=15-10=5\\text{ cm}."


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS