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MCAT Physical : Combining Lenses

Study concepts, example questions & explanations for MCAT Physical

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Example Questions

Example Question #1 : Mirrors And Lenses

In a microscope has a $\small 20cm$ tube length and objective lens with a focal length of $\small 5cm$ . The viewer's eye is $\small 25cm$ from the objective lens, and they desire a magnification of $\small -20$ . What must the focal length of the eyepiece lens be to achieve this magnification?

Possible Answers:

4cm

-5cm

-1.25cm

5cm

1.25cm

Correct answer:

5cm

Explanation:

The total magnification of a compound microscope is the product of the objective lens magnification and the eyepiece magnification.

M = m_oM_e

Objective magnification and eyepiece magnification are given by the following equations:

$m_o = -\frac{L}{f_o}$

$M_e=\frac{d_{eye}}{f_e}$

We are given the tube length and focal length of the objective lens, allowing us to solve for its magnification.

$m_o=-\frac{L}{f_o}=-\frac{20cm}{5cm}=-4$

We also know the distance of the viewer's eye. Use this value in the eyepiece magnification equation.

$M_e=\frac{d_{eye}}{f_e}=\frac{25cm}{f_e}$

Finally, combine the eyepiece magnification and objective lens magnification into the original equation for total magnification.

M=m_oM_e

$M=(-4)(\frac{25cm}{f_e})$

Use the given value for total magnification to solve for the focal length of the eyepiece lens.

$-20=(-4)(\frac{25cm}{f_e})$

$5=\frac{25cm}{f_e}$

f_e=5cm

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Example Question #11 : Mirrors And Lenses

A compound microscope consists of an eyepiece with an angular magnification of 25 and an objective lens of unknown focal length. If the length of the microscope tube is 25cm, what magnitude of objective focal length is necessary to achieve an overall magnification of 500?

Possible Answers:

0.05cm

20cm

0.8cm

{10cm}

1.25cm

Correct answer:

1.25cm

Explanation:

Relevant equations:

$m_o = -\frac{L}{f_o}$

M = m_o M_e

= length of microscope tube

f_o = focal length of objective

f_e = focal length of eyepiece

m_o = linear magnification of objective

M_e = angular magnification of eyepiece

= total magnification of microscope

Given:

L = 25cm

M_e = 25

M = 500

Step 1: Plug the expression for m_o into the equation for total magnification, .

$M =- \frac{L}{f_o}M_e$

Step 2: Rearrange to isolate the unknown, f_o

$f_o = -\frac{LM_e}{M}$

Step 3: Plug in given quantities, taking the absolute value to find the magnitude of f_o .

$f_o = |-\frac{(25cm)(25)}{500}|=1.25cm$

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Example Question #12 : Mirrors And Lenses

What diameter of telescope objective lens is necessary to resolve two stars that primarily emit 600nm light and have an angular separation of $2.44 * 10^{-8} rad$ ?

Possible Answers:

0.3m

0.04m

25m

30m

Correct answer:

30m

Explanation:

Relevant equations:

$R = \frac{1.22 \lambda}{D}$

= angular separation of sources, in radians

$\lambda$ = wavelength of light emitted by sources, in meters

= diameter of telescope

Step 1: Rearrange equation to isolate the unknown, :

$D = \frac{1.22\lambda}{R}$

Step 2: Plug in the given numbers for wavelength and angular separation:

$D = \frac{1.22(600*10^{-9}m)}{2.44*10^{-8}rad}=30m$

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MCAT Physical : Combining Lenses

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Mirrors And Lenses

Example Question #11 : Mirrors And Lenses

Example Question #12 : Mirrors And Lenses

All MCAT Physical Resources

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