Problem 69 - Arihant AIEEE Physics

Question

The refractive index of a lens material is \(\mu\) and focal length \(f\). Due to some chemical changes in the material, its refractive index has increased by \(2 \%\). The percentage decrease, in focal length for \(\mu=1.5\) will be : (a) \(4 \%\) (b) \(2 \%\) (c) \(6 \%\) (d) \(8 \%\)

Step-by-Step Solution

Answer

The percentage decrease in focal length is 6%.

Step 1: Understand the Lens Formula Relationship

The lens maker formula is given by \( \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), which connects the focal length \(f\) of the lens with the refractive index \(\mu\) and the radii of curvature \(R_1\) and \(R_2\). The percentage change in \(\mu\) will affect \(f\).

Step 2: Identify Changes in Refractive Index

The initial refractive index \(\mu = 1.5\). The refractive index increases by \(2\%\), resulting in a new refractive index \(\mu' = 1.5 \times 1.02 = 1.53\).

Step 3: Calculate Focal Length Change

Using the lens maker formula for original and new conditions:\[\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]\[\frac{1}{f'} = (1.53 - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]The difference becomes:\[\frac{1}{f'} - \frac{1}{f} = (1.53 - 1.5) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]

Step 4: Simplify and Calculate Percentage Change in Focal Length

Rearrange to find the change in focal length:\[\frac{1}{f'} - \frac{1}{f} = 0.03 \left( \frac{1}{R_1} - \frac{1}{R_2} \right)\]Assuming small percentage change:\[\frac{f'-f}{f'} \approx \frac{f-f'}{f} = \frac{0.03}{0.5} = 0.06\]This indicates a \(6\%\) decrease in the focal length.

Key Concepts

Refractive Index Change

The refractive index is a measure of how much light bends or refracts when it passes through a material. It's denoted by the symbol \( \mu \) and is essential for determining how lenses function. When light transitions from one medium to another with a different refractive index, its speed and direction change. This change is crucial for the operation of lenses in optical equipment like glasses, cameras, and microscopes.
In our exercise, the refractive index of the material initially is 1.5. However, it experiences a 2% increase due to some chemical changes, resulting in a new refractive index of 1.53. Calculating such changes is vital for understanding how modifications in material composition can affect optical properties.
For optical systems, even minor adjustments in the refractive index can significantly alter the behavior and efficiency of the lens. This highlights the importance of understanding materials and the foundational principles of optics.

Focal Length Calculation

The concept of focal length is central to optics. It's the distance between the center of a lens and its focal point, where light rays converge or diverge. In mathematical terms, the lens maker formula helps us find the focal length \( f \) of a lens: \( \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \). This equation connects the focal length with the lens's refractive index \( \mu \) and the radii of curvature \( R_1 \) and \( R_2 \).
Like in the problem we have, knowing the initial focal length allows us to predict changes when we alter the lens's refractive index. When the refractive index changes, it directly affects the lens's focal properties. It's a fundamental principle for lens design and adjustment in various fields, such as photography and vision correction.
In practice, this relationship means a small change in \( \mu \) can cause a notable change in \( f \), necessitating careful consideration in the design and handling of lenses.

Percentage Change in Optics

Understanding percent changes in optics is essential for evaluating how alterations to a lens's material impact its functionality. In our example problem, we calculate the new focal length due to a minor 2% change in the refractive index.
Using the lens maker formula, we first find the original and modified focal length equations. The difference between these helps approximate how the lens's focal length shifts. It's calculated using the relationship: \[\frac{f'-f}{f'} \approx \frac{f-f'}{f} = \frac{0.03}{0.5} = 0.06\]. From this calculation, we deduce there is a 6% decrease in the focal length.
Such calculations are critical in optical design and applications, ensuring that lens functionality remains optimal even with small material changes. Keeping track of these alterations helps engineers and scientists maintain precision in optical instruments.