Draw a ray diagram to show the working of a compound microscope.

Question

(i) Draw a ray diagram to show the working of a compound microscope. Obtain the expression for the total magnification for the final image to be formed at the near point.

(ii) In a compound microscope an object is placed at a distance of 1.5 cm from the objective of focal length 1.25 cm. If the eye-piece has a focal length of 5 cm and the final image is formed at the near point, find the magnifying power of the microscope.

Answer 1

(I)

 

Magnifying power, \(M = \frac{\beta}{\alpha}\)

\(M = \frac{tan \ \beta}{tan \ \alpha}\)

\(= \frac{A'' B ''}{C_2B''} \times \frac{C_2B''}{AB} \ [A_1B''= AB]\)

\(M = m_e \times m_0\)

\(= \left( 1+ \frac{D}{f_e}\right) \times m_0\)

\(\frac{V_0}{-u_0} \left(1+ \frac{D}{f_e}\right)\) 

(II)

u₀ = –1.5 cm

f_o = 1.25 cm

\(\frac{1}{f_0} = \frac{1}{V_0}- \frac{1}{u_0}\)

\(\frac{1}{1.25} = \frac{1}{V_0}+ \frac{1}{1.5}\) 

\(\Rightarrow \frac{1}{V_0} = \frac{1}{2.5} - \frac{1}{1.5}\)

\(= \frac{100}{125}- \frac{10}{15}\)

\(= \frac{1500 - 1250}{1875}\) 

\(\Rightarrow \frac{1}{V_0} = \frac{250}{1875}\) 

\(\Rightarrow V_0 = +7.5 \ cm\) 

\(f_e = +5 cm\) 

\(m = - \frac{V_0}{u_0} \left[1+ \frac{D}{f_e}\right]\) 

\(m = \frac{7.5}{-1.5}\left[1+ \frac{25}{5}\right]\) 

\(= \frac{-7.5}{1.5} \times 6\)

\(m = -30\)