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Home » 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? Ifwater 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?
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?
CBSE | Class 12 | NCERT | Physics | Ray Optics and Optical Instruments
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?

Answer:

We are given that

Actual depth of the needle in water here is h1=12.5cm

Apparent depth in water is h2 =9.4 cm

Refractive Index of water is given by –

\mu

The value of \mu can be calculated as follows:

\mu=\frac{h_{1}}{h_{2}}

=\frac{12.5}{9.4}

= 1.33

Therefore, 1.33 is the refractive index of water.

Water has been replaced by a liquid with a refractive index of 1.63.

The needle’s true depth remains constant, while its perceived depth fluctuates.

Let x be the needle’s new apparent depth.\mu '=\frac{h_{1}}{x}

Hence

y=\frac{h_{1}}{\mu '}

=\frac{12.5}{1.63}=7.67cm

The needle’s new apparent depth is 7.67cm. Because the value is smaller than h_{2}, the needle must be shifted back to the focus.

Distance to be moved to focus is then given as = 9.4 – 7.67

 = 1.73cm

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