Question

Find the distance of the final image from the second lens. Given \(f_1 = 10\,\text{cm}\), \(f_2 = 15\,\text{cm}\). The object is placed \(15\,\text{cm}\) to the left of the first lens and the separation between the lenses is \(15\,\text{cm}\). 

• A. \(15\,\text{cm}\)
• B. \(\infty\)
• C. \(7.5\,\text{cm}\)
• D. \(30\,\text{cm}\)

Hint:

In multiple lens problems, always solve sequentially: 1. Find image from the first lens. 2. Treat that image as the object for the second lens. 3. Apply the lens formula again.

Answer 1

The Correct Option is C

Concept: For a thin lens, the lens formula is \[ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \] where • \(u\) = object distance • \(v\) = image distance • \(f\) = focal length The image formed by the first lens acts as the object for the second lens.
Step 1: Image formed by the first lens. For first lens: \[ u = -15\,\text{cm}, \qquad f = +10\,\text{cm} \] Using lens formula \[ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \] \[ \frac{1}{v}-\frac{1}{-15}=\frac{1}{10} \] \[ \frac{1}{v}+\frac{1}{15}=\frac{1}{10} \] \[ \frac{1}{v}=\frac{1}{10}-\frac{1}{15} \] \[ \frac{1}{v}=\frac{3-2}{30} \] \[ \frac{1}{v}=\frac{1}{30} \] \[ v=30\,\text{cm} \] Thus the first image is \(30\,\text{cm}\) to the right of the first lens.
Step 2: Object distance for the second lens. Distance between lenses \(=15\,\text{cm}\). Hence the image formed by the first lens lies \[ 30-15=15\,\text{cm} \] to the right of the second lens. Thus for the second lens \[ u=+15\,\text{cm} \] \[ f=+15\,\text{cm} \]
Step 3: Image formed by the second lens. Using lens formula \[ \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \] \[ \frac{1}{v}-\frac{1}{15}=\frac{1}{15} \] \[ \frac{1}{v}=\frac{1}{15}+\frac{1}{15} \] \[ \frac{1}{v}=\frac{2}{15} \] \[ v=\frac{15}{2} \] \[ v=7.5\,\text{cm} \] Thus the final image is \[ \boxed{7.5\,\text{cm}} \] to the right of the second lens.