Question

In a compound microscope, an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. The eyepiece has a focal length of 5 cm. The final image is formed at infinity. Calculate the distance between the objective and the eyepiece.

Hint:

In a compound microscope, the distance between the objective and eyepiece is crucial for proper image formation at infinity. The image formed by the objective lens serves as the object for the eyepiece, and the distance is adjusted accordingly to form a final image at infinity.

Answer 1

Given data: - Object distance for objective \( u_{\text{obj}} = -1.5 \, \text{cm} \) (since the object is on the left side of the lens) - Focal length of objective \( f_{\text{obj}} = 1.25 \, \text{cm} \) - Focal length of eyepiece \( f_{\text{eyepiece}} = 5 \, \text{cm} \) - The final image is formed at infinity. 1. Find the image distance for the objective lens \( v_{\text{obj}} \): Using the lens formula for the objective lens: \[ \frac{1}{f_{\text{obj}}} = \frac{1}{v_{\text{obj}}} - \frac{1}{u_{\text{obj}}} \] Substituting the known values: \[ \frac{1}{1.25} = \frac{1}{v_{\text{obj}}} - \frac{1}{-1.5} \] \[ \frac{1}{v_{\text{obj}}} = \frac{1}{1.25} + \frac{1}{1.5} = 0.8 + 0.6667 = 1.4667 \] \[ v_{\text{obj}} = \frac{1}{1.4667} \approx 0.682 \, \text{cm} \] So, the image formed by the objective lens is at a distance of approximately 0.682 cm. 2. Find the object distance for the eyepiece \( u_{\text{eyepiece}} \): The object for the eyepiece is the image formed by the objective lens. This image acts as the object for the eyepiece, so: \[ u_{\text{eyepiece}} = -v_{\text{obj}} = -0.682 \, \text{cm} \] 3. Find the distance between the objective and eyepiece: The image formed by the objective lens is at a distance of \( 0.682 \, \text{cm} \) from the objective. Since the final image is formed at infinity, the object distance for the eyepiece must be such that the image is formed at infinity. This happens when the object for the eyepiece is placed at the focal point of the eyepiece. The distance between the objective and the eyepiece is the sum of the image distance for the objective and the focal length of the eyepiece: \[ d = v_{\text{obj}} + f_{\text{eyepiece}} = 0.682 + 5 = 5.682 \, \text{cm} \] Thus, the distance between the objective and the eyepiece is approximately 5.68 cm.