Question

Explain about compound microscope and obtain the equation for magnification. Compound microscope:

Answer 1

The lens near the object, called the objective, forms a real, inverted, magnified image of the object. This serves as the object for the second lens which is the eyepiece. Eyepiece serves as a simple microscope that produces finally an enlarged and virtual image. The first inverted image formed by the objective is to be adjusted close to, but within the focal plane of the eyepiece so that the final image is formed nearly at infinity or at the near point. The final image is inverted with respect to the original object. We can obtain the magnification for a compound microscope.

Magnification of compound microscope : 

From the ray diagram, the linear magnification due to the objective is,

m₀ = \(\frac{h'}{f}\)

From the figure, tan β = \(\frac{h}{f_0}\) = \(\frac{h'}{L'}\), then

\(\frac{h'}{h}\) = \(\frac{L}{f_0}\); m₀= \(\frac{L}{f_0}\)

Here, the distance L is between the first focal point of the eyepiece to the second focal point of the objective. This is called the tube length L of the microscope as f₀ and f_e are comparatively smaller than L. If the final image is formed at P (near point focusing), the magnification m_e of the eyepiece is,

m_e= 1 + \(\frac{D}{F_e}\)

The total magnification m in near point focusing is,

m = m₀m_e = \((\frac{L}{F_0})\) \((1 + \frac{D}{f_e})\)

If the final image is formed at infinity (normal focusing), the magnification me of the eyepiece is,

m_e= \(\frac{D}{f_e}\)

The total magnification m in normal focusing is,

m = m₀m_e = \((\frac{L}{f_0})\) \((\frac{D}{f_e})\)