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Chapter 26: Q29PE (page 958)

You switch from an 1.40NA 60×oil immersion objective to an 1.40NA 60×oil immersion objective. What are the acceptance angles for each? Compare and comment on the values. Which would you use first to locate the target area on your specimen?

Short Answer

Expert verified

The acceptance angle isθ=135∘59'25"

Step by step solution

01

Concept Introduction

Oil immersion is a technique for improving a microscope's resolving power. This is accomplished by immersing both the objective lens and the specimen in a transparent oil with a high refractive index, which increases the objective lens' numerical aperture.

02

Calculating the angles

The job appears to have been conducted wrongly because the numerical apertures are equal. in the case of oil. The acceptance angle may be calculated using the formula below:

NA=nsinθ2

From here, we have:

θ=2arcsinNAn=2arcsin1.41.51=135∘59'25"

If we had two targets to choose from, the one with the highest acceptance angle would be the first to be used.

Therefore, the required solution isθ=135∘59'25".

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Most popular questions from this chapter

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

What was the previous far point of a patient who had laser vision correction that reduced the power of her eye by 7.00 D, producing normal distant vision for her?

An amoeba is 0.305 cm away from the 0.300 cm focal length objective lens of a microscope. (a) Where is the image formed by the objective lens? (b) What is this image’s magnification? (c) An eyepiece with a 2.00 cm focal length is placed 20.0 cm from the objective. Where is the final image? (d) What magnification is produced by the eyepiece? (e) What is the overall magnification? (See Figure 26.16.)

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

Calculate the power of the eye when viewing an object 3.00 maway.

Consider a telescope of the type used by Galileo, having a convex objective and a concave eyepiece as illustrated in Figure 26.23(a). Construct a problem in which you calculate the location and size of the image produced. Among the things to be considered are the focal lengths of the lenses and their relative placements as well as the size and location of the object. Verify that the angular magnification is greater than one. That is, the angle subtended at the eye by the image is greater than the angle subtended by the object.

The contact lens prescription for a nearsighted person is -4.00 D and the person has a far point of 22.5 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?
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