Physics, asked by ThakurRajSingh24, 9 months ago

{\boxed{\red{\underline{\sf{QUESTION :}}}}}
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?
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Answers

Answered by Sanayasilawat
18

Explanation:

η=

9.4

12.5

=1.329

When the water is replaced by liquid of 1.63

η=

apparentdepth

12.5

1.63=

x

12.5

x=7.67

So Distance by which microscope have to be moved is 9.4−7.67=1.73cm

Apparent depth = real depth / refractive index

or

9.4 = 12.5 / μ

μ = 12.5 /9.4

μ = 1.33.

Now for the other part.

Apparent depth h' = 12.5 / μ'

h' = 12.5 /1.63

h' = 7.67

Previously apparent depth was 9.4 cm

Hence Microscope have to be moved by

= (9.4 - 7.67 )

= 1.73 cm

hope it's help u

Answered by Anonymous
95

Given :

First case :

▪ Medium : water

▪ Real depth = 12.5cm

▪ Apparent depth = 9.4cm

Second case :

▪ Medium : liquid

▪ Real depth = 12.5cm

▪ Refractive index of liquid = 1.63

To Find :

↗ Refractive index of water. (First case)

↗ Distance by which microscope have to be moved to focus on needle. (Second case)

Formula :

☞ Relation between refractive index of medium, real depth and apparent depth is given by

\bigstar\:\underline{\boxed{\bf{\red{n=\dfrac{Real\:depth}{Apparent\:depth}}}}}

CalculaTion :

First case :

\Rightarrow\sf\:n_w=\dfrac{H_r}{H_a}\\ \\ \Rightarrow\sf\:n_w=\dfrac{12.5}{9.4}\\ \\ \Rightarrow\underline{\boxed{\bf{\blue{n_w=1.33}}}}\:\gray{\bigstar}

Second case :

\dashrightarrow\sf\:n_l=\dfrac{H_r}{H_a}\\ \\ \dashrightarrow\sf\:1.63=\dfrac{12.5}{H_a}\\ \\ \dashrightarrow\sf\:H_a=\dfrac{12.5}{1.63}\\ \\ \dashrightarrow\bf\:H_a=7.67\:cm

☣ So, Distance by which microscope have to be moves is given by

\implies\sf\:x=9.4-7.67\\ \\ \implies\underline{\boxed{\bf{\green{x=1.73\:cm}}}}\:\gray{\bigstar}

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