Conversation started Feb 29, 2020 at 9:30.
Guru Vishnu
Guru Vishnu
Feb 29, 2020 09:30
@JohnRennie: Hi sir :-)
John Rennie
John Rennie
@GuruVishnu hi
Guru Vishnu
Guru Vishnu
in The h Bar, 2 mins ago, by Guru Vishnu
As per the new Cartesian sign convention in optics, is it ok to determine the nature of object or image based on the signs of object and image distance?
Or in other words; is positive value of something always virtual and negative value always real for object and the exact opposite for images, sir?
John Rennie
John Rennie
Yes, the whole point of a sign convention is that it enables you to work out the details of the system you are studying.
Guru Vishnu
Guru Vishnu
+ object distance - virtual object
- object distance - real object
+ image distance - real image
- image distance - virtual image
John Rennie
John Rennie
@GuruVishnu That's true for lenses. It gets a bit more complicated for mirrors.
Guru Vishnu
Guru Vishnu
Feb 29, 2020 09:33
@JohnRennie Ok sir. So the above condition always holds good irrespective of the optical device- lens or mirror or anything else. Am I right, or are there any other exceptions to this rule?
@JohnRennie Yes sir. Just now realised it for concave mirrors even though negative image distance gives a real image and not a virtual as suggested by this.
Is it also applicable in all circumstances for a spherical interface separating two optical media, sir?
Like this:
Medium 1 . . . . . . . . ( . . . . . . . . Medium 2
John Rennie
John Rennie
Yes
moved
Guru Vishnu
Guru Vishnu
Feb 29, 2020 09:50
DCP-14-442-17
Two refracting media are separated by a spherical interface as shown in the figure:
$\mu_2$ . . . . . . . . ( . . . . . . . . $\mu_1$
(a) If $\mu_2>\mu_1$, then there cannot be a real image of real object
(b) If $\mu_2>\mu_1$, then there cannot be a real image of virtual object
(c) If $\mu_1>\mu_2$, then there cannot be a virtual image of virtual object
(d) If $\mu_1>\mu_2$, then there cannot be a real image of real object
My approach:
Using the lens maker's formula, I determined the possible signs the image distance $v$ can take for a particular sign of the object distance $u…
@JohnRennie: Hi sir. If you're free, could you please reply to the message above? Thank you.
John Rennie
John Rennie
The way I remember how to handle a curved interface is that it's like a plano lens.
So in this case you effectively have a planoconvex lens.
Guru Vishnu
Guru Vishnu
Ok sir. For calculation purposes, I think it's safe to assume a planoconvex lens as a simple biconvex lens. Is this permitted sir?
John Rennie
John Rennie
I'm not sure that makes it any easier ...
Guru Vishnu
Guru Vishnu
As for both planoconvex and bi-convex lens, the focal length for both sides remains the same. We're just neglecting spherical aberration.
John Rennie
John Rennie
Let me draw a quick diagram:
user image
Guru Vishnu
Guru Vishnu
Feb 29, 2020 10:03
Ok sir. For virtual object the object distance is positive. And also the image distance is positive as the focal length is positive.
John Rennie
John Rennie
@GuruVishnu OK there's my diagram.
The focal length of the lens is going to be:
Guru Vishnu
Guru Vishnu
@JohnRennie One small issue: I think $R_2=\infty$
John Rennie
John Rennie
$$ \frac{1}{f} = \left(\frac{n_1}{n_2} - 1\right) \frac{1}{R_1} $$
Guru Vishnu
Guru Vishnu
@JohnRennie Yes sir.
John Rennie
John Rennie
@GuruVishnu oops yes.
In (a) $n_2 > n_1$ so $f < 0$ i.e. it is a diverging lens. And for a real object a diverging lens always produces a virtual object. So (a) is true.
Guru Vishnu
Guru Vishnu
Feb 29, 2020 10:06
@JohnRennie Yes sir. No problem with (a). The (c) is the one giving me some trouble.
John Rennie
John Rennie
OK. In (c) $n_1 > n_2$ so it is a converging lens.
Guru Vishnu
Guru Vishnu
@JohnRennie Ok sir.
John Rennie
John Rennie
The lens equation is:
$$ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} $$
Guru Vishnu
Guru Vishnu
For both $f$ and $u$ is positive, $v$ is always positive.
Then how can a virtual image be impossible?
It's always the possibility.
John Rennie
John Rennie
For a virtual image $v < 0$ and for a virtual object $u > 0$, and we know $f > 0$.
So let's write our equation with these signs:
$$ \frac{-1}{|v|} = \frac{+1}{|f|} + \frac{+1}{|u|} $$
OK so far?
Guru Vishnu
Guru Vishnu
Feb 29, 2020 10:10
@JohnRennie Completely understood my fault. I considered +ve sign for virtual image instead of negative sign. Thank you very much sir :-)
John Rennie
John Rennie
:-)
 
Conversation ended Feb 29, 2020 at 10:10.