Optics: Applying Sign Conventions Twice?

[Nikhil_kumar]

Please provide me with some help in optics. This doubt is in relation to the use of sign conventions in optics. Whenever we prove anything in optics, say for example, when we prove the mirror formula or the lens formula or the lens-maker's formula, we apply the sign conventions in the derivation of the proof itself (u=-ve, f=+ve or -ve etc., according to the New Cartesian Conventions). Then while solving problems based on these formulae, why do we again have to apply the sign conventions according to the data given in the question? I mean, to solve problems based on the lens formula , the mirror formula etc. why do we have to apply the conventions twice? After all the conventions have already been applied during the course of proof itself.

For eg, The lens formula: 1/f=1/v - 1/u is derived in case of real image by convex lens by putting u=-ve, f=+ve v=+ve during the course of proof itself.

 

[rl.bhat]

In lens or mirror we get two types of images. The formula is derived for both. While solving the problems, we have apply the sign convention again to take into account the nature of the image.

 

[ogb p]

Then while solving a problem, if the object distance u is given to be +ve, we again put u=+ve. Why is this necessary?

I can understand your confusion regarding the use of sign conventions in optics. The reason for applying sign conventions twice is to ensure consistency in the calculations and to avoid any errors.

Firstly, let's understand the purpose of using sign conventions in optics. These conventions are used to determine the direction and sign of the object distance (u), image distance (v), and focal length (f) in relation to the lens or mirror. This helps in simplifying the calculations and also provides a standard way of representing these values.

Now, when we are deriving formulas such as the lens formula or the mirror formula, we are using the sign conventions to determine the relative signs of these distances. This is essential in proving the formula and understanding the underlying principles of optics.

However, when we are solving problems using these formulas, we need to apply the sign conventions again to the given data in order to get the correct numerical values. This is because the values of u, v, and f may vary depending on the specific problem and it is important to use the correct signs to get accurate results.

In the example you have mentioned, if the object distance (u) is given to be +ve, it means that the object is located on the same side as the incident light, which is different from the assumption made during the proof where u was taken to be -ve for a real image produced by a convex lens. Therefore, to get the correct value of v, we need to use the correct sign for u in the lens formula.

In conclusion, applying sign conventions twice is necessary to ensure consistency and accuracy in the calculations. It may seem redundant, but it is an important step in solving problems in optics. I hope this explanation helps in clarifying your doubt.