Question

A microscope has a 2.2 −cm -focal-length eyepiece and a 0.50 −cm objective. Assuming a relaxed normal eye.

Calculate the position of the object if the distance between the lenses is 16.2 cm .

Calculate the total magnification.

Answer 1

Answer 2

1. Calculate the position of the object ($u_o$$u_o$) :-

For the eyepiece (used as a simple magnifier in relaxed mode):

$$v_e=\mathrm{\infty }⟹u_e=f_e=2.2\text{cm}$$

The image formed by the objective ($v_o$) is at the focal point of the eyepiece:

$$v_o=L-u_e=16.2\text{cm}-2.2\text{cm}=14.0\text{cm}$$

Using the lens formula for the objective:

$$\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o}$$$$\frac{1}{0.50}=\frac{1}{14.0}-\frac{1}{u_o}$$$$2=0.0714-\frac{1}{u_o}$$$$\frac{1}{u_o}=0.0714-2=-1.9286$$$$u_o=-\frac{1}{1.9286}\approx -0.52\text{cm}$$

The object is placed 0.52 cm to the left of the objective lens.

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2. Calculate the total magnification ($M$$M$) :-

For a microscope with the final image at infinity:

$$M=M_o\times M_e$$

where:

• $M_o$ is the magnification of the objective:

  $$M_o=\frac{v_o}{u_o}=\frac{14.0}{0.52}\approx 26.92$$

• $M_e$ is the magnification of the eyepiece:

  $$M_e=\frac{D}{f_e}=\frac{25\text{cm}}{2.2\text{cm}}\approx 11.36$$

Total magnification:

$$M=26.92\times 11.36\approx 305.7$$

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 Answers is :-

1. The object is placed 0.52 cm from the objective lens.

2. The total magnification is 306