Question 141: Ray Optics

Question

A vessel of depth '$$d$$' is half filled with oil of refractive index $$n_{1}$$ and the other half is filled with water of refractive index $$n_{2}$$. The apparent depth of this vessel when viewed from above will be-

Accepted Answer

To find the apparent depth of the vessel when viewed from above, we can calculate the apparent depths of the oil and water separately and then add them together.
The formula to find the apparent depth ($$h_{apparent}$$) is:
$$h_{apparent} = \frac{h_{real}}{n}$$
Where $$h_{real}$$ is the actual depth, and $$n$$ is the refractive index of the medium.
For the oil (with depth $$\frac{d}{2}$$ and refractive index $$n_1$$):
$$h_{oil} = \frac{\frac{d}{2}}{n_{1}}$$
For the water (with depth $$\frac{d}{2}$$ and refractive index $$n_2$$):
$$h_{water} = \frac{\frac{d}{2}}{n_{2}}$$
Now, add the two apparent depths together to find the total apparent depth:
$$h_{total} = h_{oil} + h_{water} = \frac{\frac{d}{2}}{n_{1}} + \frac{\frac{d}{2}}{n_{2}}$$
Combine the terms:
$$h_{total} = \frac{d}{2}\left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right)$$
Now, find a common denominator for the fractions:
$$h_{total} = \frac{d}{2}\left(\frac{n_{1} + n_{2}}{n_{1}n_{2}}\right)$$
Multiply the fractions:
$$h_{total} = \frac{d(n_{1} + n_{2})}{2n_{1}n_{2}}$$
The correct answer is:
$$\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}$$

Answer

$$\frac{2 d\left(n_{1}+n_{2}\right)}{n_{1} n_{2}}$$

Answer

$$\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}$$

Answer

$$\frac{d n_{1} n_{2}}{2\left(n_{1}+n_{2}\right)}$$

Answer

$$\frac{d n_{1} n_{2}}{\left(n_{1}+n_{2}\right)}$$

Ray Optics

A vessel of depth '$d' is half filled with oil of refractive index n_{1} and the other half is filled with water of refractive index n_{2}$. The apparent depth of this vessel when viewed from above will be-

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