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Chapter 24: Problem 39

The wing of an insect is \(1.0 \mathrm{mm}\) long. When viewed through a microscope, the image is \(1.0 \mathrm{m}\) long and is located \(5.0 \mathrm{m}\) away. Determine the angular magnification.

Short Answer

Expert verified
Answer: The angular magnification of the microscope is 50.

Step by step solution

01

Calculate the angle subtended by the insect wing when viewed with the naked eye

To calculate the angle subtended by the wing in the naked eye, we assume standard viewing distance of 25 cm or 0.25 m. Then we use the small angle approximation formula: tan θ ≈ θ = object size (height) / distance, where θ is the angle subtended by the object. θ_naked_eye = (1.0 mm) / (0.25 m) We need to convert the object size in mm to meters: 1.0 mm = 0.001 m. θ_naked_eye = (0.001 m) / (0.25 m)
02

Calculate the angle subtended by the insect wing when viewed through the microscope

This time, we're given the size of the image when viewed through the microscope (1.0 m), and the distance of the image from the eye (5.0 m). Using the small angle approximation formula again, we can calculate the angle subtended by the image formed through the microscope. θ_microscope = object size (height) / distance θ_microscope = (1.0 m) / (5.0 m)
03

Calculate the angular magnification

The angular magnification is the ratio of the angle subtended by the object when viewed through a microscope to the angle subtended by the object when viewed with the naked eye. M = θ_microscope / θ_naked_eye M = (1.0 m / 5.0 m) / (0.001 m / 0.25 m)
04

End: Final Answer

With all the necessary calculations, we can now determine the angular magnification of the microscope. M = (1.0/5.0)/(0.001/0.25) = 0.2/0.004 = 50. So the angular magnification of the microscope is 50.

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Most popular questions from this chapter

A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
Show that if two thin lenses are close together \((s,\) the distance between the lenses, is negligibly small), the two lenses can be replaced by a single equivalent lens with focal length \(f_{\mathrm{eq}} .\) Find the value of \(f_{\mathrm{eq}}\) in terms of \(f_{1}\) and \(f_{2}.\)
Suppose the distance from the lens system of the eye (cornea + lens) to the retina is \(18 \mathrm{mm}\). (a) What must the power of the lens be when looking at distant objects? (b) What must the power of the lens be when looking at an object \(20.0 \mathrm{cm}\) from the eye? (c) Suppose that the eye is farsighted; the person cannot see clearly objects that are closer than $1.0 \mathrm{m}$. Find the power of the contact lens you would prescribe so that objects as close as \(20.0 \mathrm{cm}\) can be seen clearly.
Repeat Problem \(40(\mathrm{c})\) using a different eyepiece that gives an angular magnification of 5.00 for a final image at the viewer's near point \((25.0 \mathrm{cm})\) instead of at infinity.
A camera has a telephoto lens of 240 -mm focal length. The lens can be moved in and out a distance of \(16 \mathrm{mm}\) from the film plane by rotating the lens barrel. If the lens can focus objects at infinity, what is the closest object distance that can be focused?
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