9.17 (a) Figure 9.32 shows a cross-section of a ‘light pipe’ made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figure.
(b) What is the answer if there is no outer covering of the pipe?
Refractive index of outer covering of the pipe, = 1.44
Given:-
Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’
The refractive index of the inner core-outer core interface, is given as
=
or =
i’ = 59
For the critical angle, total reflection take place only when i > i’ i.e. i > 59
Maximum angle of reflection, = 90 = 31
Let be the maximum angle of incidence and be the maximum angle of reflection.
or
1.6
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9.17 Refractive index of the glass fibre, = 1.68
Refractive index of outer covering of the pipe, = 1.44
Given:-
Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’
The refractive index of the inner core-outer core interface, is given as
=
or =
i’ = 59
For the critical angle, total reflection take place only when i > i’ i.e. i > 59
Maximum angle of reflection, = 90 = 31
Let be the maximum angle of incidence and be the maximum angle of reflection.
or
1.68
The entire rays incident at angles lying in the range of 0 < i < 60 will suffer total internal reflection.
If the outer covering is absent, then:
Refractive index of the outer pipe, = Refractive index of air = 1
For the angle of incidence, i = 90 , we can write Snell’s law at ‘air – pipe’ interface as :
= = 1.68
= =
r = 36.53
i’ = 90 36.53 = 53.47
Since i’ > r, all incident rays will suffer total internal reflection.
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<p><strong>9.17 </strong>Refractive index of the glass fibre, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span> = 1.68</p><p>Refractive index of outer covering of the pipe, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math></span> = 1.44</p><p>Given:-</p><p>Angle of incidence = i, Angle of refraction = r, Angle of incidence at the interface = i’</p><p>The refractive index of the inner core-outer core interface, <span title="Click to copy mathml"><math><mi>μ</mi></math></span> is given as</p><p><span title="Click to copy mathml"><math><mi>μ</mi><mo>=</mo><mi></mi><mfrac><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow></mfrac></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mi>i</mi><mi>'</mi></mrow></mrow></mrow></mrow></mfrac></math></span></p><p><span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mi>i</mi><mi>'</mi></mrow></mrow></mrow></mrow></mfrac><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>1.44</mn></mrow></mrow><mrow><mrow><mn>1.68</mn></mrow></mrow></mfrac></math></span> or <span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mi>i</mi><mi>'</mi></mrow></mrow></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1.44</mn></mrow></mrow><mrow><mrow><mn>1.68</mn></mrow></mrow></mfrac></math></span></p><p>i’ = 59 <span title="Click to copy mathml"><math><mo>°</mo></math></span></p><p>For the critical angle, total reflection take place only when i > i’ i.e. i > 59 <span title="Click to copy mathml"><math><mo>°</mo></math></span></p><p>Maximum angle of reflection, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>τ</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></math></span> = 90 <span title="Click to copy mathml"><math><mo>°</mo><mo>-</mo><mi>i</mi><mi>'</mi></math></span> = 31 <span title="Click to copy mathml"><math><mo>°</mo></math></span></p><p>Let <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>i</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></math></span> be the maximum angle of incidence and <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></math></span> be the maximum angle of reflection.</p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>=</mo><mi></mi><mfrac><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><msub><mrow><mrow><mi>i</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow></mrow></mrow><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><msub><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow></mrow></mrow></mfrac></math></span> or <span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><msub><mrow><mrow><mi>i</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow><mo>=</mo><mi></mi><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mi></mi><mo>×</mo><mi></mi><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><msub><mrow><mrow><mi>r</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow></math></span></p><p><span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><msub><mrow><mrow><mi>i</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub></mrow></mrow><mo>=</mo><mi></mi></math></span> 1.68 <span title="Click to copy mathml"><math><mo>×</mo><mi>s</mi><mi>i</mi><mi>n</mi><mn>31</mn><mo>°</mo></math></span></p><p><span title="Click to copy mathml"><math><msub><mrow><mrow><mi>i</mi></mrow></mrow><mrow><mrow><mi>m</mi><mi>a</mi><mi>x</mi></mrow></mrow></msub><mo>=</mo><mn>59.91</mn><mo>°</mo></math></span> <span title="Click to copy mathml"><math><mo>≈</mo><mn>60</mn><mo>°</mo></math></span></p><p>The entire rays incident at angles lying in the range of 0 < i < 60 will suffer total internal reflection.</p><p>If the outer covering is absent, then:</p><p>Refractive index of the outer pipe, <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math></span> = Refractive index of air = 1</p><p>For the angle of incidence, i = 90 <span title="Click to copy mathml"><math><mo>°</mo></math></span> , we can write Snell’s law at ‘air – pipe’ interface as :</p><p><span title="Click to copy mathml"><math><mfrac><mrow><mrow><mi>s</mi><mi>i</mi><mi>n</mi><mi></mi><mi>i</mi></mrow></mrow><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mi>r</mi></mrow></mrow></mrow></mrow></mfrac></math></span> = <span title="Click to copy mathml"><math><msub><mrow><mrow><mi>μ</mi></mrow></mrow><mrow><mrow><mn>2</mn><mi></mi></mrow></mrow></msub></math></span> = 1.68</p><p><span title="Click to copy mathml"><math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mi>r</mi><mo>=</mo><mi></mi></mrow></mrow></math></span> <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mi>i</mi></mrow></mrow></mrow></mrow><mrow><mrow><mn>1.68</mn></mrow></mrow></mfrac></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mn>90</mn><mo>°</mo></mrow></mrow></mrow></mrow><mrow><mrow><mn>1.68</mn></mrow></mrow></mfrac></math></span> = <span title="Click to copy mathml"><math><mfrac><mrow><mrow><mn>1</mn></mrow></mrow><mrow><mrow><mn>1.68</mn></mrow></mrow></mfrac></math></span></p><p>r = 36.53 <span title="Click to copy mathml"><math><mo>°</mo></math></span></p><p>i’ = 90 <span title="Click to copy mathml"><math><mo>°</mo><mo>-</mo><mi></mi></math></span> 36.53 <span title="Click to copy mathml"><math><mo>°</mo></math></span> = 53.47 <span title="Click to copy mathml"><math><mo>°</mo></math></span></p><p>Since i’ > r, all incident rays will suffer total internal reflection.</p>
A total refractive prism is also known as a total internal reflection prism. It is an optical prism that is designed for reflecting 100% of the incident light. This happens since this prism uses the principle of total internal reflection. These prisms are oriented and shaped in a specific way so that the light that enters at a specific angle is completely reflected inside the prism. A right-angle prism, porro prism, dove prism and roof prism are some of the examples of total reflective prism.
Total deviation in a prism is the total angle by which the light ray gets bent as it passes through the prism. It is an angle between incident ray and emergent ray of the prism. When a light enters the prism, it will bend towards the normal. After that, it will travel through the prism and bend away from the normal as it exits. Total deviation is the sum of these two from which the apex angle is subtracted.
The formula for total deviation for a prism is as follows:
There are different types of glasses that are used in optical instruments, including the following:
Crown glass (K): This glass is used in eyeglasses, microscopes and cameras. It is used in prisms and windows in optical systems. Crown glass has a low refractive index, low dispersion and excellent transparency in visible spectrum.
Flint Glass (F): This glass, when combined with crown glass, can correct chromatic aberration in lenses. They are also used in prisms for spectroscopy.
Extra-low dispersion glass: These glasses are used in premium optics that are also used for making high-quality camera lenses, telescopes and binoculars.
Optical instruments can have some of the following defects that may impact their performance, which have arisen due to design limitations, manufacturing and physical properties of light:
Chromatic Aberration: This defect occurs because of the different wavelengths of light that refract at slightly different angles when they pass through the lens. It causes them to focus on different points.
Spherical Aberration: This happens because light rays pass through the edges of spherical lens or reflect off spherical mirror focus at different point than rays that pass through the center.
Astigmatism: This type of defect occurs due to the uneven cu
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Optical instruments can have some of the following defects that may impact their performance, which have arisen due to design limitations, manufacturing and physical properties of light:
Chromatic Aberration: This defect occurs because of the different wavelengths of light that refract at slightly different angles when they pass through the lens. It causes them to focus on different points.
Spherical Aberration: This happens because light rays pass through the edges of spherical lens or reflect off spherical mirror focus at different point than rays that pass through the center.
Astigmatism: This type of defect occurs due to the uneven curvature of lenses or mirrors, which causes light to focus differently in horizontal and vertical planes.
Field Curvature: One of the defects in optical instruments is field curvature, which occurs due to flat image sensors and film that cannot perfectly match the curved focal plane of a lens.
Yes, optical instruments are used in modern medicine for many purposes including surgery, monitoring, research and diagnosis. Let us take a look at each one by one:
Many optical instruments are used for visualizing internal structures for diagnosis of a disease and its monitoring. These include Ophthalmoscope, Endoscope, Colposcope and Dermatoscope.
Optical instruments are also used for precision and minimally invasive surgeries, including Laparoscope, Arthroscope and Surgical Microscopes.
Lasers are used for cutting, therapy and coagulation since they have precision and minimal invasiveness. CO? Laser, Excimer Laser and Fiber Optic
...more
Yes, optical instruments are used in modern medicine for many purposes including surgery, monitoring, research and diagnosis. Let us take a look at each one by one:
Many optical instruments are used for visualizing internal structures for diagnosis of a disease and its monitoring. These include Ophthalmoscope, Endoscope, Colposcope and Dermatoscope.
Optical instruments are also used for precision and minimally invasive surgeries, including Laparoscope, Arthroscope and Surgical Microscopes.
Lasers are used for cutting, therapy and coagulation since they have precision and minimal invasiveness. CO? Laser, Excimer Laser and Fiber Optic Lasers are some of the optical instruments.
Optical instruments also help in monitoring vital signs in the body as well as for analysing biological samples. Pulse Oximeter, Spectrophotometer and Optical Coherence Tomography (OCT) are some of the optical instruments.
For cellular-level analysis and medical research, optical instruments like the Confocal Microscope and Fluorescence Microscope are used.
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