# Refraction through a glass slab + reflection off a concave mirror

1. Apr 8, 2015

### Suraj M

1. The problem statement, all variables and given/known data
An object is placed 21 cm in front of a concave mirror of radius of curvature of 10 cm. A glass slab of thickness 3 cm and RI =1.5; is then placed close to the mirror. Find the position of the final image.
Take distance of closer side of block as 1 cm from the mirror.

2. Relevant equations
$\frac{1}{f}= \frac{1}{u} + \frac{1}{v}$
u is the object dist. and v is the image dist.
$lateral~ shift~=~t\frac{\sin(i-r)}{\cos(r)}$

3. The attempt at a solution
Ignoring the block i get $v = -6.56 cm$
I don't know how to consider the block... i tried this..

i need the length of B'B'₂
I have the measure of a perpendicular drawn from A' to the line through A'₂ which was suppose to go through the focus..

Last edited: Apr 8, 2015
2. Apr 8, 2015

### Tanya Sharma

First consider a point object on the principal axis for sake of simplicity . Second ,forget about the mirror for a while . The ray from object undergoes twice the refraction from the slab .

Can you calculate the normal shift produced by the slab ?

3. Apr 8, 2015

### Suraj M

But how, doesn't that depend on the angle of incidence.
Which ray?
passing through the centre of curvature →normal shift = 0 as it is refracted twice ..

4. Apr 8, 2015

### Tanya Sharma

Forget the mirror.

I suggested you to consider a point object placed at __ cm from the slab . Draw any ray from the object at any angle from the priniple axis . Angle is not important .Center of curvature is not involved here .

Last edited: Apr 8, 2015
5. Apr 8, 2015

### Suraj M

$lateral ~shift = t \frac{\sin(i-r)}{\cos(r)}$
so the lateral shift does depend on the incident angle, doesn't it?
but as you said if i take the ray at an angle($=i$) to the principal axis then
lateral shift = $3\sin(i) - 4.5 \cos(i)$

6. Apr 8, 2015

### Tanya Sharma

Normal shift and lateral shift are two different things . We are interested in calculating normal shift not lateral . I have no idea about the formula you have mentioned . Please check your notes on how to calculate the normal shift produced by the glass slab .Again , normal shift doesn't depend on the angle .

Have a look at the picture attached .

You need to calculate distance OA .

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• ###### shift.PNG
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Last edited: Apr 8, 2015
7. Apr 8, 2015

### Suraj M

I thought we needed lateral shift.. ok normal shift ..then its just 5.67 cm ($\frac{d}{n}$)
but i still think its lateral shift and not normal shift, normal shift would be this..
I think the diagram you have drawn is actually for lateral shift as the emergent ray is parallel to the incident ray...
So i cannot apply the formula $d' = \frac{d}{n}$or normal shift = $d(1-\frac{1}{n})$ where n is RI of glass wrt air

8. Apr 8, 2015

### Tanya Sharma

Don't you think the emergent ray would be parallel to the incident ray after two refractions due to the glass slab ?

This is the right formula . So , now what is the normal shift ?

However what is the answer given to the problem?

9. Apr 9, 2015

### Suraj M

The answer given is 7.67 cm.
Yes it will be, so if the emergent ray is parallel to the incident ray... there is a lateral shift, not normal shift, right?? A component of the lateral shift is the distance between the position and apparent position.

10. Apr 9, 2015

### Tanya Sharma

Ok...I am getting this value .

Calculate the value of normal shift . Then , find the apparent distance of the object from the mirror . Now , use the mirror formula .

11. Apr 9, 2015

### Suraj M

wait should i do this??
$d' = \frac{17}{1.5}$ then $d' + 3 + 1 = u$??
then i get 7.42 cm
is that right?

12. Apr 9, 2015

### Tanya Sharma

Sorry , I don't understand your working .

1) What value do you get for normal shift ?
2) What is the apparent distance of the object from the mirror ?

13. Apr 9, 2015

### Suraj M

for (1) i wanted to know if i should take the distance between the glass block(17 cm ) or the distance from the mirror..
anyway i get $d' = 11.33 cm$ and apparent object distance as 15.33 cm

14. Apr 9, 2015

### Tanya Sharma

This is incorrect .

Normal shift = $(1-\frac{1}{\mu})t$ , where $\mu$ is the refractive index and $t$ is the thickness of the glass slab .

Now please answer the two questions I asked in post#12 .

15. Apr 9, 2015

### Suraj M

ok is the normal shift 5.66 cm ??
if so then apparent distance is 17 - 5.66 = 11.33 cm
which is the same as $\frac{17}{n}$

16. Apr 9, 2015

### Tanya Sharma

No

No

I have given you the formula in post #14 . Please use it .

17. Apr 11, 2015

### Suraj M

I wasn't sure what you were trying to do here, but i think i get the idea..
you're trying to calculate the effective object distance which would become $(u=21 -1 = 20 cm )$
so then applying mirror formula i get v = 6.67 cm ... but due to the slab there is a second refraction causing the image to shift by 1 cm ... = 6.67+1= 7.67 cm
I'm sorry it took me so much time.. :) and for irritating you so much...
Thanks a ton!

18. Jun 12, 2016

### Anikash Chakraborty

What if we place the glass slab at 5 cm from the mirror? Will the image be still formed at 7.67 cm? Or will there be some change. In my opinion there shouldn't be any but please can anyone confirm?

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