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QJEEM23Ph01c2: JEE Main 2023 Physics - Class 12 JEEM23Ph01c2 Solution: A vessel of depth ' ' is half filled with oil of refractive…

A vessel of depth ' dd ' is half filled with oil of refractive index n1n_{1} and the other half is filled with water of refractive index n2n_{2} . The apparent depth of this vessel when viewed from above will be-

Multiple Choice Options

  • A 2d(n1+n2)n1n2\frac{2 d\left(n_{1}+n_{2}\right)}{n_{1} n_{2}}
  • B d(n1+n2)2n1n2\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}
  • C dn1n22(n1+n2)\frac{d n_{1} n_{2}}{2\left(n_{1}+n_{2}\right)}
  • D dn1n2(n1+n2)\frac{d n_{1} n_{2}}{\left(n_{1}+n_{2}\right)}

Step-by-Step Verified Solution

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 ( happarenth_{apparent} ) is:

happarent=hrealnh_{apparent} = \frac{h_{real}}{n}

Where hrealh_{real} is the actual depth, and nn is the refractive index of the medium.

For the oil (with depth d2\frac{d}{2} and refractive index n1n_1 ):

hoil=d2n1h_{oil} = \frac{\frac{d}{2}}{n_{1}}

For the water (with depth d2\frac{d}{2} and refractive index n2n_2 ):

hwater=d2n2h_{water} = \frac{\frac{d}{2}}{n_{2}}

Now, add the two apparent depths together to find the total apparent depth:

htotal=hoil+hwater=d2n1+d2n2h_{total} = h_{oil} + h_{water} = \frac{\frac{d}{2}}{n_{1}} + \frac{\frac{d}{2}}{n_{2}}

Combine the terms:

htotal=d2(1n1+1n2)h_{total} = \frac{d}{2}\left(\frac{1}{n_{1}} + \frac{1}{n_{2}}\right)

Now, find a common denominator for the fractions:

htotal=d2(n1+n2n1n2)h_{total} = \frac{d}{2}\left(\frac{n_{1} + n_{2}}{n_{1}n_{2}}\right)

Multiply the fractions:

htotal=d(n1+n2)2n1n2h_{total} = \frac{d(n_{1} + n_{2})}{2n_{1}n_{2}}

The correct answer is:

d(n1+n2)2n1n2\frac{d\left(n_{1}+n_{2}\right)}{2 n_{1} n_{2}}

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