A microscope is focused on a mark on a table, when the mark is covered by a plate of glass 2m thick, the…

Analysis

This is a classic optics problem involving apparent depth. When a transparent medium (like a glass plate) is placed over an object, the object appears to be raised. The microscope must be moved upward to compensate for this shift so that it can focus on the "apparent" position of the mark.

Concept

The formula for the apparent shift ($s$) caused by a glass plate of thickness ($t$) and refractive index ($n$) is:

Where:

• $s$ = shift (the distance the microscope is raised) = $0.67\text{ cm}$
• $t$ = real thickness of the glass = $2\text{ cm}$
• $n$ = refractive index of the glass

Solution

1. Convert units: Ensure all units are consistent. Here, both $s$ and $t$ are in centimeters, so we can proceed directly.
2. Set up the equation: $$0.67 = 2 \left(1 - \frac{1}{n}\right)$$
3. Solve for $n$: Divide both sides by 2: $$\frac{0.67}{2} = 1 - \frac{1}{n}$$ $$0.335 = 1 - \frac{1}{n}$$
4. Rearrange to isolate $1/n$: $$\frac{1}{n} = 1 - 0.335$$ $$\frac{1}{n} = 0.665$$
5. Calculate $n$: $$n = \frac{1}{0.665} \approx 1.5037$$

Rounding to two decimal places, we get 1.50.

Conclusion

Comparing our result to the options provided, the correct value is 1.50.

Exam Tip

When dealing with "apparent depth" or "shift" problems, always remember that the shift is the difference between the real depth and the apparent depth. If you ever forget the formula, just remember: Shift = Real Depth - Apparent Depth, where Apparent Depth = $\frac{\text{Real Depth}}{n}$.