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Standard VIII

Physics

Apparent Depth

What is the l...

Question

What is the length of the image of the rod in mirror, according to the observer in air? (refractive index of the liquid is $μ$ ).

μ L + L

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L + \frac{L}{μ}

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L μ + \frac{L}{μ}

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N o n e o f t h e s e

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Solution

The correct option is A $μ L + L$
$\begin{matrix} The rod in air is seen as same length seen directly = L The rod in water is seen as \begin{matrix} \frac{μ}{1} = \frac{Apparat depth}{L} Apparent depth = L μ so total length = L + μ L \end{matrix} so option A is correct. \end{matrix}$

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What is the length of the image of the rod in mirror, according to the observer in air? (refractive index of the liquid is μ).

The correct option is A μL+L The rod in air is seen as same length seen directly = L The rod in water is seen as μ1= Apparat depth L Apparent depth =Lμ so total length =L+μL so option A is correct.

What is the length of the image of the rod in mirror, according to the observer in air? (refractive index of the liquid is $μ$ ).

The correct option is A $μ L + L$
$\begin{matrix} The rod in air is seen as same length seen directly = L The rod in water is seen as \begin{matrix} \frac{μ}{1} = \frac{Apparat depth}{L} Apparent depth = L μ so total length = L + μ L \end{matrix} so option A is correct. \end{matrix}$