A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
1. Given Data
- Real Depth (h1): 12.5 cm
- Apparent Depth (h2): 9.4 cm
- Refractive Index of Liquid (nl): 1.63
2. Calculating Refractive Index of Water
The refractive index (n) is defined as the ratio of Real Depth to Apparent Depth.
n = Real Depth / Apparent Depth
Substituting the values:
- n = 12.5 / 9.4
- n ≈ 1.33
3. Calculation for New Liquid
Now, the water is replaced by a liquid with refractive index 1.63. The real depth remains the same (12.5 cm).
We calculate the new Apparent Depth (h'):
h' = Real Depth / Refractive Index
h' = 12.5 / 1.63
h' ≈ 7.67 cm
4. Microscope Displacement
To find how much the microscope needs to move, we find the difference between the two apparent depths.
Shift = Initial Apparent Depth - New Apparent Depth
Shift = 9.4 cm - 7.67 cm
Shift = 1.73 cm
Final Answer
- Refractive Index of Water: 1.33
- Microscope Movement: 1.73 cm
- Direction: Upwards (away from the tank).
Why move the microscope up?
The new liquid has a higher refractive index (1.63 > 1.33), which causes the light to bend more. This makes the image of the needle appear shallower (closer to the surface) at 7.67 cm depth compared to 9.4 cm. Therefore, the microscope must be raised by 1.73 cm to focus on the new, higher image position.
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