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A glass slab of known thickness t = 2 cm is placed on a horizontal platform. A traveling microscope is set up in such a way that it views the image of a distant object through the glass slab. The microscope is focused on the image without the glass slab. When the glass slab is placed, the microscope needs to be moved vertically upward by h = 0.5 cm to focus on the image again. Determine the refractive index n of the glass slab.

Option: 1

3\cdot \sin i


Option: 2

4\cdot \sin i


Option: 3

2\cdot \sin i


Option: 4

3\cdot \sin i


Answers (1)

best_answer

Step 1: The situation involves the glass slab acting as a medium with a certain refractive index n through which light passes.

Step 2: Consider the setup. When the microscope is focused on the distant object without the glass slab, the light travels through air (with refractive index n_{\text {air }}=1).

Step 3: When the glass slab is placed on the platform, light travels through the glass slab (with refractive index n ) and then through air.

Step 4: By Snell's Law, we have:
            n_{\text {air }} \sin i=n \sin r
where i is the angle of incidence and r is the angle of refraction inside the glass slab.
Step 5: Since n_{\text {air }}=1, we get:

            \sin i=n \sin r

Step 6: When the microscope is moved vertically upward by h units, the light ray gets displaced by h as well. The triangle formed by the displacement h, thickness of the glass slab t, and the angle of refraction r is a right-angled triangle.
Step 7: Using trigonometry, we have:
            \tan r=\frac{h}{t}
Step 8: Substitute the value of \tan r in terms of h and t into the equation from step 5:
            \sin i=n \cdot \frac{h}{t}
Step 9: Solve for n :
            n=\frac{t \cdot \sin i}{h}=\frac{2 \mathrm{~cm} \cdot \sin i}{0.5 \mathrm{~cm}}=4 \cdot \sin i
So, the refractive index (n) of the glass slab is 4 \cdot \sin i.
So, option B is correct.

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