Question

An object is placed at a distance of 30 cm from a convex lens of focal length 100 cm when it moves 5 cm towards the lens how much is the image being shifted?

Hint:

Always strictly adhere to the Cartesian sign convention. Remember that for a virtual object moving closer to the focal point from the optical center, the virtual image also moves closer to the optical center.

Answer 1

Step 1: Understanding the Concept:
This problem requires applying the thin lens formula to find the image positions for two different object distances. The shift in the image is the absolute difference between these two image positions.
Step 2: Key Formula or Approach:
The thin lens formula is:
\[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \]
Where:
\(v\) = image distance
\(u\) = object distance
\(f\) = focal length
Sign convention: For a convex lens, \(f\) is positive. Object distance \(u\) is generally taken as negative.
Step 3: Detailed Explanation:
Initial case:
Focal length, \(f = +100\) cm.
Initial object distance, \(u_1 = -30\) cm.
Let the initial image distance be \(v_1\).
\[ \frac{1}{v_1} - \frac{1}{u_1} = \frac{1}{f} \]
\[ \frac{1}{v_1} - \left(\frac{1}{-30}\right) = \frac{1}{100} \]
\[ \frac{1}{v_1} + \frac{1}{30} = \frac{1}{100} \]
\[ \frac{1}{v_1} = \frac{1}{100} - \frac{1}{30} \]
\[ \frac{1}{v_1} = \frac{3 - 10}{300} = \frac{-7}{300} \]
\[ v_1 = -\frac{300}{7} \text{ cm} \approx -42.86 \text{ cm} \]
(The negative sign indicates a virtual image formed on the same side as the object).
Final case:
The object moves 5 cm towards the lens.
New object distance, \(u_2 = -(30 - 5) = -25\) cm.
Let the new image distance be \(v_2\).
\[ \frac{1}{v_2} - \left(\frac{1}{-25}\right) = \frac{1}{100} \]
\[ \frac{1}{v_2} + \frac{1}{25} = \frac{1}{100} \]
\[ \frac{1}{v_2} = \frac{1}{100} - \frac{1}{25} \]
\[ \frac{1}{v_2} = \frac{1 - 4}{100} = \frac{-3}{100} \]
\[ v_2 = -\frac{100}{3} \text{ cm} \approx -33.33 \text{ cm} \]
Shift in image:
The shift is the magnitude of the difference between the two image positions.
\[ \text{Shift} = |v_2 - v_1| = \left| -\frac{100}{3} - \left(-\frac{300}{7}\right) \right| \]
\[ \text{Shift} = \left| -\frac{100}{3} + \frac{300}{7} \right| \]
\[ \text{Shift} = \left| \frac{-700 + 900}{21} \right| \]
\[ \text{Shift} = \frac{200}{21} \text{ cm} \approx 9.52 \text{ cm} \]
Step 4: Final Answer:
The image shifts by \(\frac{200}{21}\) cm towards the lens.