A breaker is filled with a liquid to a height of 14cm. The apparent depth of a needle fixed at the bottom of the beaker is measured to be 10cm by a microscope. Whats if the refractive index of liquid? The height of the liquid in the beaker is now raised to 21cm
Answers
Given:
A breaker is filled with a liquid to a height of 14cm.
The apparent depth of a needle fixed at the bottom of the beaker is measured to be 10cm by a microscope.
To find:
Whats if the refractive index of liquid? The height of the liquid in the beaker is now raised to 21cm
Solution:
From given, we have,
Actual depth of the needle in liquid (h1) = 14 cm
Apparent depth of the needle in liquid (h2) = 10 cm
Refractive index of liquid = μ
The value of μ can be calculated as,
μ = h1/ h2 = 14/10 = 1.4
Hence, the refractive index of the liquid is about 1.4.
The height of the liquid in the beaker is now raised to 21 cm
The value of h2 (apparent depth) can be calculated as,
μ = h1/ h2
1.4 = 21/h2
h2 = 15 cm
The refractive index of the liquid is 1.4 and the value of apparent depth is 15 cm
Explanation:
refractive index of the liquid = real depth/apparent depth
formula
real depth/apparent depth=refractive index
1st part
real=14 apparently it's 10 (as per the que)
therefore
n=14/10=1.4(refractive index )
2nd part
here real depth changes to 21cm
and the refractive index we got as 1.4
therefore apparent depth by the formula would be
15 cm
now the distance moved by the microscope to focus on the needle again is
apparent depth before - apparent depth after changing the real depth
15 -10 cm = 5 cm
the second part is according to question given in the book
hope it helped
best of luck!!
mark as brainliest