while deriving lens we use sign convention and so we get the formula 1/f=1/v-1/u

then while using this formula why do we use sign convention again to solve any problem?

When we derive the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), we indeed start with a specific sign convention. This convention helps us define the directions of the object distance (u), image distance (v), and focal length (f). However, when we apply this formula to solve problems, we must consistently use the sign convention again to ensure that our calculations yield accurate results. Let’s break this down further.

The Importance of Sign Convention

Sign conventions are crucial in optics because they provide a standardized way to interpret distances and directions. In the case of lenses, the common sign convention is as follows:

• Object distance (u): Positive if the object is on the same side as the incoming light (real object), negative if on the opposite side (virtual object).
• Image distance (v): Positive if the image is on the opposite side of the lens from the object (real image), negative if on the same side (virtual image).
• Focal length (f): Positive for converging lenses and negative for diverging lenses.

Why Use Sign Convention Again?

When you apply the lens formula to specific problems, using the sign convention again ensures that you maintain consistency in your calculations. Here’s why this is essential:

• Directionality: The sign convention helps clarify the direction of light rays and the position of the object and image. If you don’t adhere to it, you might end up with incorrect signs for distances, leading to erroneous results.
• Mathematical Consistency: The lens formula is derived based on the assumption of a particular sign convention. If you change the signs arbitrarily while solving a problem, you disrupt the mathematical relationships established during the derivation.
• Real vs. Virtual Images: The sign convention helps differentiate between real and virtual images. For instance, if you calculate a negative value for v, it indicates a virtual image, which is crucial for understanding the nature of the image formed.

Example to Illustrate

Let’s consider a practical example. Suppose you have a converging lens with a focal length of +10 cm. An object is placed 30 cm in front of the lens. To find the image distance (v), you would first identify the object distance (u) as -30 cm (since it’s a real object). Now, applying the lens formula:

Using the formula:

\( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Substituting the values:

\( \frac{1}{10} = \frac{1}{v} - \frac{1}{(-30)} \)

Now, solving for v:

\( \frac{1}{10} = \frac{1}{v} + \frac{1}{30} \)

To combine the fractions, find a common denominator:

\( \frac{3}{30} = \frac{1}{v} + \frac{1}{30} \)

Thus:

\( \frac{3}{30} - \frac{1}{30} = \frac{1}{v} \)

\( \frac{2}{30} = \frac{1}{v} \)

So, \( v = 15 \) cm, which is positive, indicating a real image formed on the opposite side of the lens.

Final Thoughts

In summary, using the sign convention consistently throughout your calculations is vital for obtaining correct results in lens problems. It ensures clarity in the relationships between object distance, image distance, and focal length, allowing you to accurately interpret the nature of the images formed by lenses. By adhering to these conventions, you can confidently tackle a variety of optical problems.