Question

Optics - Spherical Surface:
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Answer 1

I need help with this one. @Mashy

Answer 2

what grade are you in

Answer 3

\[
\frac{\mu_1}{u} + \frac{\mu_2}{v} = \frac{\mu_2 - \mu_1}{r}
\]

Answer 4

Oh, I am past "grades" :) My nephew in India is preparing for a college entrance examination and he sends me math and physics problems that he got stuck in. I tried to answer as much as I can but when I get stuck I post them here hoping someone would help.

Answer 5

if i gave you my username and password could get my smartscore up quick

Answer 6

It will happen on its own over a period of time as you go about helping others.

Answer 7

i know but i want it up quick can you do it for me please

Answer 8

Sorry, I can't.

Answer 9

If I didn't mess up the signs, upon first refraction, the object forms a virtual image on the same side as the object at v = 700 cm. Not sure what to do after that.

Answer 10

Ok.. when it comes to optics.. and spherical surfaces, we need sign conventions.. now there are two.. depending on which ur Nephew's syllabus is following, the formula changes a wee bit

if i use the international and more standard cartesian sign conventions then it is \[\frac{\mu2}{v} - \frac{\mu1}{u} = \frac{\mu2 -\mu1}{R}\]

mu2 -- R.I. of refracted medium (here its the spherical surface), v -- image distance
mu1 -- R.I of incident medium (here, its the air), u-- object distance
and R - radius of curvature

The distances are always measured from the pole.. and incident direction is positive

if he is following CBSE syllabus.. then this is the one

sometimes there s another .. not very uncommon system .. the gaussian system.. in that we use real objects and images +ve and virtual ones negative.. so that sues ur formula

Answer 11

good job @Mashy :P

Answer 12

Figured out this problem. I am getting K = 3 but it would be nice if this can be confirmed.

Good to know about the two different sign conventions. Thanks @mashy. The only Physics book I have for reference is Halliday-Resnick-Walker and they adopt the Gaussian system. With that I get v = -700 cm after the very FIRST refraction from the spherical surface. Therefore, it is a virtual image formed on the same side as the object at a distance of 700 cm from the spherical surface. This virtual image is at a distance of 700 + 100 = 800 cm from the plane mirror which upon reflection will form a virtual image 800 cm "behind" the mirror. The distance of this virtual object from the spherical surface is 900 cm. And by applying the above equation again, for the SECOND refraction at the spherical surface, I get v = +90 cm. which is on the same side as the original object. 90 = 30 K; K = 3.

Yes, my nephew is following the CBSE syllabus and so I will use the other sign convention when I email him the solution.