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Chapter 35: Q. 21 (page 1017)

A narrow beam of white light is incident on a sheet of quartz. The beam disperses in the quartz, with red light (l≈400nm)traveling at an angle of 26.3°with respect to the normal and violet light (l≈400nm) traveling at 25.7° . The index of refraction of quartz for red light is 1.45. What is the index of refraction of quartz for violet light?

Short Answer

Expert verified

The index of refraction of quartz for violet light is1.48.

Step by step solution

01

Given Information.

We have given that:

Red light = (l≈700nm),

An angel = 26.3°,

Normal violet light (l≈400nm),

Traveling 25.7°,

Index of refraction 1.45.

We need find the index of refraction of quartz for violet light.

02

Simply.

As we know:

λ1≈700nm

θ1=26.3°

n2=1.45

λ2≈400nm

θ2=25.7°

03

Calculation

We need to find n2by using formula:

n1sinθ1=n2sinθ2

1.00sinθ1=1.45sin26.3°

39.97°=θ1

⇒1.00sin39.97°=n2sin25.7°
0.6423=n20.4336
n2=0.64230.4336

n2=1.48

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Most popular questions from this chapter

A camera takes a properly exposed photo at f/5.6and 1125s. What shutter speed should be used if the lens is changed to f/4.0?

In FIGURE P35.30, what are the position, height, and orientation of the final image? Give the position as a distance to the right or left of the lens.

Two converging lenses with focal lengths of 40 cm and 20 cm are 10 cm apart. A 2.0-cm-tall object is 15 cm in front of the 40-cm-focal-length lens. a). Use ray tracing to find the position and height of the image. Do this accurately using a ruler or paper with a grid, then make measurements on your diagram.

b). Calculate the image position and height. Compare with your ray-tracing answers in part a.

The resolution of a digital camera is limited by two factors:

diffraction by the lens, a limit of any optical system, and the fact

that the sensor is divided into discrete pixels. Consider a typical

point-and-shoot camera that has a 20-mm-focal-length lens and

a sensor with 2.5@mm@wide pixels.

a. First,ass ume an ideal, diffractionless lens. At a distance of

100 m, what is the smallest distance, in cm, between two

point sources of light that the camera can barely resolve? In

answering this question, consider what has to happen on the

sensor to show two image points rather than one. You can use

s′ = f because s W f.

b. You can achieve the pixel-limited resolution of part a only if

the diffraction width of each image point is no greater than

1 pixel in diameter. For what lens diameter is the minimum

spot size equal to the width of a pixel? Use 600 nm for the

wavelength of light.

c. What is the f-number of the lens for the diameter you found in

part b? Your answer is a quite realistic value of the f-number

at which a camera transitions from being pixel limited to

being diffraction limited. For f-numbers smaller than this

(larger-diameter apertures), the resolution is limited by the

pixel size and does not change as you change the aperture. For

f-numbers larger than this (smaller-diameter apertures), the

resolution is limited by diffraction, and it gets worse as you

“stop down” to smaller apertures

The resolution of a digital cameras is limited by two factors diffraction by the lens, a limit of any optical system, and the fact that the sensor is divided into discrete pixels. consirer a typical point-and--shoot camera that has a 20-mm-focal-lengthlens and a sensor with 2.5-μm-widepixels.

(a) . First, assume an ideal, diffractionless lens, at a distance of 100m,what is the smallest distance, in cmbetween two point sources of light that the camera can barely resolve? in answering this question, consider what has to happen on the sensor to show two image points rather than one you can use S1=fbecauses>>f.

(b) . You can achieve the pixel-limied resolution of part a only if the diffraction which of each image point no greater than the diffraction width of image point is no greater than 1pixel in diameter. for what lens diameter is the minimum spot size equal to the width of a pixel ? use 600nmfor the wavelength of light.

(c). what is the f-numberof the lens for the diameter you found in part b? your answer is a quite realistic value of the f-numberat which a camera transitions from being pixel limited to being diffraction limited for f-numbersmaller than this (larger-diameter apertures), the resolution is limited by the pixel size and does not change as you change the apertures. for f-numberlarger than this (smaller-diameter apertures). the resolution is limited by diffraction and it gets worse as you "stop down" to smaller apertures.
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