Question

A microscope is focused on a mark, then a glass slab of refractive index 1.5 and thickness of 6 cm is placed on the mark to get the mark again in focus, the microscope should be moved

• A. 4 cm
• B. 2 cm
• C. 6 cm
• D. 8 cm

Answer 1

Answer: (B)

2 cm

Answer 2

Here's a step-by-step solution:

1. Understand the concept of apparent depth and normal shift:
   When an object is viewed through a denser medium (like glass) from a rarer medium (like air), it appears to be at a shallower depth than its actual (real) depth. This phenomenon is due to refraction. The relationship between real depth ($d$) and apparent depth ($d'$) is given by: $d' = \frac{d}{\mu}$
   where $\mu$ is the refractive index of the denser medium.
   The apparent shift (or normal shift) ($S$) is the difference between the real depth and the apparent depth:
   $S = d - d' = d - \frac{d}{\mu} = d \left(1 - \frac{1}{\mu}\right)$

2. Identify the given values:
   Thickness of the glass slab (real depth, $d$) = 6 cm
   Refractive index of the glass slab ($\mu$) = 1.5

3. Calculate the apparent depth:
   $d' = \frac{6 \text{ cm}}{1.5} = 4 \text{ cm}$
   This means that when viewed from above, the mark appears to be only 4 cm below the top surface of the glass slab.

4. Calculate the normal shift:
   The mark was originally at a depth of 6 cm (relative to the top surface if the slab were already there). After placing the slab, it appears to be at a depth of 4 cm.
   The apparent position of the mark has shifted upwards (closer to the observer/microscope).
   Shift ($S$) = Real depth - Apparent depth
   $S = 6 \text{ cm} - 4 \text{ cm} = 2 \text{ cm}$
   Alternatively, using the formula:
   $S = d \left(1 - \frac{1}{\mu}\right) = 6 \text{ cm} \left(1 - \frac{1}{1.5}\right) = 6 \text{ cm} \left(1 - \frac{2}{3}\right) = 6 \text{ cm} \left(\frac{1}{3}\right) = 2 \text{ cm}$

5. Determine the direction of microscope movement:
   Initially, the microscope is focused on the actual mark. When the glass slab is placed, the mark appears to be raised by 2 cm. This means the apparent image of the mark is 2 cm closer to the microscope objective than the actual mark was. To bring the mark back into focus, the microscope objective needs to be moved closer to this apparent image. Therefore, the microscope must be moved downwards by 2 cm.

The question asks for the magnitude of the movement.