Answer:
The focal length of eye piece is 6.52 cm.
Explanation:
Given that,
Angular Magnification of the microscope M = -46
the distance between the lens in microscope L= 16 cm
The focal length of objective f₀ = 1.5 cm
Normal near point N = 25 cm
Have to find focal length of eye piece f ₙ =?
The angular magnification is given by
M ≈ - (L-fₙ)N/f₀fₙ
Rearranging for fₙ
fₙ =L(1 - Mf₀/N)⁺¹
=18/2.76
fₙ = 6.52 cm
The focal length of eye piece is 6.52 cm.
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Answer:
The position on the x axis is 0.32 m.
Explanation:
Given that,
Point charge = 27 nC
Charge = 6 nC
Distance = 1
We need to calculate the distance
Using formula of electric field
Put the value into the formula
Hence, The position on the x axis is 0.32 m.
If it's volume changes when you move it to the new container it would be a solid
