Refraction through a glass slab + reflection off a concave mirror

[Suraj M]

Homework Statement

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.

Homework Equations

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

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..

 

[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 ?

 

[Suraj M]

But how, doesn't that depend on the angle of incidence.

The ray from object undergoes twice t

Which ray?
passing through the centre of curvature →normal shift = 0 as it is refracted twice ..

 

[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 .

 

[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 = $3\sin(i) - 4.5 \cos(i)$

 

[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 .

 

[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

 

[Tanya Sharma]

I think the diagram you have drawn is actually for lateral shift as the emergent ray is parallel to the incident ray...

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

normal shift = $d(1-\frac{1}{n})$ where n is RI of glass wrt air

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

However what is the answer given to the problem?

 

[Suraj M]

The answer given is 7.67 cm.

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

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.

 

[Tanya Sharma]

The answer given is 7.67 cm.

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 .

 

[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?

 

[Tanya Sharma]

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

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 ?

 

[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

 

[Tanya Sharma]

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

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 .

 

[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 $\frac{17}{n}$

 

[Tanya Sharma]

ok is the normal shift 5.66 cm ??

No

if so then apparent distance is 17 - 5.66 = 11.33 cm

No

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

 

[Suraj M]

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

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!

 

[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?

 

[MARAUDER18]

please i want the position of the 2 images formed in the central problem .i can get only one. Mam Tanya sharma please help

 

[MARAUDER18]

Plz anyone