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MCAT Physical : Thin Lens Equation

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Example Questions

Example Question #4 : Mirrors And Lenses

A virtual image is formed $\small 8cm$ from a convex mirror with a focal length of $\small 10cm$ . How far from the mirror is the object that created this image?

Possible Answers:

4.4cm

40cm

22.5cm

2cm

2.5cm

Correct answer:

40cm

Explanation:

Use the equation:

$\frac{1}{f}=\frac{1}{d_i}+\frac{1}{d_o}$

Focal length is negative for convex mirrors, and image distance is negative for virtual images. We are given these values in the question, allowing us to calculate the object distance.

$\frac{1}{-10}=\frac{1}{-8}+\frac{1}{d_o}$

$\frac{1}{d_o}=\frac{1}{40}$

$d_{o}= 40cm$

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Example Question #5 : Mirrors And Lenses

How far from a converging lens must an object be placed to produce an image that is NOT real and inverted? Given the answer as d_o in terms of the focal length, .

Possible Answers:

d_o >2f

d_o=2f

d_o=f

f<d_o<2f

Correct answer:

d_o=f

Explanation:

When an object is placed a distance from a converging lens or mirror that is equal to the focal length, no image is produced. To test this out, stand in front of a single concave mirror and continue to back up until you no longer see an image. Once you've reached this point, you will be standing one focal length away from the mirror.

When the object is less than one focal length away from the converging lens/mirror, the image will be virtual and upright.

If you don't have these trends committed to memory, you can derive them from the equation $\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$ .

When d_i is a negative integer the image is virtual, and when it is a positive integer the image is real.

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Example Question #1 : Thin Lens Equation

A certain farsighted person cannot focus on objects closer to his eyes than $\small 140cm$ . What focal length eyeglass lenses are needed in order to focus on a newspaper held at $\small 35cm$ from the person's eyes, if the glasses are worn $\small 2cm$ from his eyes?

Possible Answers:

28cm

43cm

27cm

47cm

2cm

Correct answer:

43cm

Explanation:

First we need the object and image distances away from the eyeglass lenses. Here, the newspaper is the object and the $\small 140cm$ focal point is where the image needs to be located. Find the distance from the object to the lens, and the distance of the image to the lens, by subtracting out the distance from the lens to the eye.

$d_{ol}=d_{oe}-d_{le}=35cm-2cm=33cm$

$d_{il}=d_{ie}-d_{le}=140cm-2cm=138cm$

Now apply the thin lens equation to determine focal length.

$\frac{1}{f} = \frac{1}{d_{i}}+\frac{1}{d_{o}}$

Recall that if the image is on the same side of the lens as the object, then image distance is negative.

$\frac{1}{f}=\frac{1}{-138}+\frac{1}{33}\approx0.023$

$f = \frac{1}{0.023}\approx 43cm$

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Example Question #2 : Thin Lens Equation

An object is placed 50cm in front of a concave mirror of radius 60cm. How far from the mirror is the image?

Possible Answers:

75cm

60cm

20cm

300cm

120cm

Correct answer:

75cm

Explanation:

Relevant equations:

r = 2f

$\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$

Step 1: Find the focal length of the mirror.

$r = 2f \rightarrow f = \frac{r}{2}=\frac{60cm}{2}=30cm$

Step 2: Plug this focal length and the object distance into the lens equation.

$\frac{1}{30}=\frac{1}{50}+\frac{1}{d_i}$

d_i = 75cm

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Example Question #8 : Mirrors And Lenses

An object that is 3cm tall is placed 30cm from a convex spherical mirror of radius 40cm, along its central axis. What is the height of the image that is formed?

Possible Answers:

7.5cm

6.0cm

1.2cm

12.0cm

1.7cm

Correct answer:

1.2cm

Explanation:

Relevant equations:

r = 2f

$\frac{1}{f}= \frac{1}{d_i}+ \frac{1}{d_o}$

$M = -\frac{d_i}{d_o}= \frac{h_i}{h_o}$

Step 1: Find the focal length of the mirror (remembering that convex mirrors have negative focal lengths, by convention).

$r = 2f \rightarrow f = \frac{r}{2}= \frac{40cm}{2}=20cm \rightarrow -20cm$

Step 2: Find the image distance using the thin lens equation.

$\frac{1}{-20}=\frac{1}{30}+\frac{1}{d_i}$

d_i = -12cm

Step 3: Use the magnification equation to relate the object distances and heights.

$-\left (\frac{-12cm}{30cm} \right )= \frac{h_i}{3cm}$

h_i = 1.2cm

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MCAT Physical : Thin Lens Equation

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #4 : Mirrors And Lenses

Example Question #5 : Mirrors And Lenses

Example Question #1 : Thin Lens Equation

Example Question #2 : Thin Lens Equation

Example Question #8 : Mirrors And Lenses

All MCAT Physical Resources

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