Question

Find the distance of the image from object O, formed by the combination of lenses in the figure : 

 

• A. 10 cm
• B. 20 cm
• C. 75 cm
• D. infinity

Hint:

For multi-lens systems, be systematic. Calculate the image for the first lens, then use its position to find the object distance for the second lens. Pay close attention to the signs and relative positions. A positive image distance means the image is real and on the opposite side of the lens from the object. This image acts as an object for the next lens. If it falls beyond the next lens, it's a virtual object (positive object distance).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We have a system of three lenses. We need to find the position of the final image formed by this combination and then calculate its distance from the original object.
Step 2: Key Formula or Approach:
We will apply the lens formula, \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\), sequentially for each lens. The image formed by one lens serves as the object for the next lens. We use the standard sign convention (light travels from left to right).
Step 3: Detailed Explanation:
For the first lens (L1, convex):
Focal length, \(f_1 = +10\) cm.
Object distance, \(u_1 = -30\) cm.
Using the lens formula: \[ \frac{1}{v_1} - \frac{1}{-30} = \frac{1}{10} \] \[ \frac{1}{v_1} = \frac{1}{10} - \frac{1}{30} = \frac{3 - 1}{30} = \frac{2}{30} = \frac{1}{15} \] So, \(v_1 = +15\) cm. The image I1 is formed 15 cm to the right of L1.
For the second lens (L2, concave):
Focal length, \(f_2 = -10\) cm.
The distance between L1 and L2 is 5 cm. The image I1 is 15 cm from L1. This means I1 is \(15 - 5 = 10\) cm to the right of L2. Since it is on the right side, it acts as a virtual object for L2.
Object distance for L2, \(u_2 = +10\) cm.
Using the lens formula: \[ \frac{1}{v_2} - \frac{1}{+10} = \frac{1}{-10} \] \[ \frac{1}{v_2} = -\frac{1}{10} + \frac{1}{10} = 0 \] So, \(v_2 = \infty\). The rays leaving L2 are parallel to the principal axis. The image I2 is formed at infinity.
For the third lens (L3, convex):
Focal length, \(f_3 = +30\) cm.
The object for L3 is the image I2, which is at infinity. This means parallel rays are incident on L3.
When the object is at infinity (\(u_3 = \infty\)), the image is formed at the focal point of the lens.
Image distance for L3, \(v_3 = f_3 = +30\) cm.
The final image I3 is formed 30 cm to the right of L3.
Calculating the final distance:
The original object O is 30 cm to the left of L1.
The final image I3 is 30 cm to the right of L3.
The total distance between O and I3 is the sum of the distances along the axis.
Distance = (Distance of O from L1) + (Distance between L1 and L3) + (Distance of I3 from L3)
Distance between L1 and L3 = 5 cm + 10 cm = 15 cm.
Total distance = 30 cm + 15 cm + 30 cm = 75 cm.
Alternatively, let L1 be at origin (x=0). Object O is at x = -30 cm. L1 is at x = 0 cm. L2 is at x = 5 cm. L3 is at x = 15 cm. Final image I3 is 30 cm to the right of L3, so its position is \(x = 15 + 30 = 45\) cm. Distance between object O and image I3 = \(x_{I3} - x_O = 45 - (-30) = 75\) cm.
Step 4: Final Answer:
The distance of the final image from the object O is 75 cm.