Optics - Find the refractive index

[Saitama]

The formula you used to find the image distance is derived from Snell's Law, using small-angle approximation. Was not it shown to you?

Yes, it was shown to me, I forgot about it. Thanks a lot ehild! :)

By the way, you get the same refractive index with your method if you solve the last equation correctly.

I still don't get the correct answer with that equation. I always end up with 0=-4. I feel there is some sign error in my last equation but I can't spot it.

 

[ehild]

\frac{1}{R}-\cfrac{\mu}{\cfrac{(4-\mu)R}{\mu-2}}=\frac{1-\mu}{-R}\rightarrow
1-\frac{\mu(\mu-2)}{4-\mu}=-1+\mu\rightarrow 2-\mu+\frac{\mu(2-\mu)}{4-\mu}=0 \rightarrow \left(2-\mu\right) \left(1+\frac{\mu}{4-\mu}\right)=0

2-μ can be factored out, and you get a product equal to zero. One of the factors must be zero. The second one can not. Never "simplify" an equation dividing by a term that contains the unknown!

 

[Saitama]

\frac{1}{R}-\cfrac{\mu}{\cfrac{(4-\mu)R}{\mu-2}}=\frac{1-\mu}{-R}\rightarrow
1-\frac{\mu(\mu-2)}{4-\mu}=-1+\mu\rightarrow 2-\mu+\frac{\mu(2-\mu)}{4-\mu}=0 \rightarrow \left(2-\mu\right) \left(1+\frac{\mu}{4-\mu}\right)=0

2-μ can be factored out, and you get a product equal to zero. One of the factors must be zero. The second one can not. Never "simplify" an equation dividing by a term that contains the unknown!

Thanks a lot ehild! I need to be more careful while solving the equations. Thanks. :)