The absolute refractive index of benzene is 1.50 and that of water is 1.33. Determine the refractive index of benzene relative to water. 1.991.131.531.03 If the refractive index of diamond with respective air is 2.5, what will be the refractive index of air with respective diamond?2.50.50.40.8 Ultraviolet light with a wavelength of 300 nm passes from air into a substance with a refractive index of 1.5. What is the wavelength of light inside substance in AoAo ? (Given that 1𝑛𝑚=10Ao1nm=10 Ao )2300Ao2300 Ao 2100Ao2100 Ao 2000Ao2000 Ao None of the above Light travelling through a certain optical fiber at a speed of 1.95×108𝑚/𝑠1.95×10 8 m/s enters a different fiber where its speed changes to 2.30×108𝑚/𝑠2.30×10 8 m/s. If the refractive index of the first fiber is 1.48, determine the refractive index of the second fiber relative to the first. Correct up to 2 decimal points.1.180.850.120.02 Light travels through a 5 mm thick sheet of transparent plastic with a refractive index of 1.20. Calculate the time taken by the light to pass through the plastic sheet and emerge outside from other side. (Given: Speed of light in vacuum = 3×108𝑚/𝑠3×10 8 m/s)3×10−8𝑠3×10 −8 s2×10−8𝑠2×10 −8 s3×10−11𝑠3×10 −11 s2×10−11𝑠2×10 −11 s Light from a laser pointer strikes a plane mirror at an angle of 30o 30 o to the normal. If the angle of incidence decreases by 8o 8 o , what is the new angle of reflection?30o30 o 38o 38 o 22o 22 o 45o 45 o A 3.9 cm tall toy is placed at a distance of 24 cm from a concave mirror. If the image of toy formed is one-third the size of the toy and is inverted, determine the height of the image and focal length of the mirror.-1.3 cm and +6 cm respectively-1.3 cm and -6 cm respectively-1.3 cm and -12 cm respectively-1.3 cm and +12 cm respectively
Question
The absolute refractive index of benzene is 1.50 and that of water is 1.33. Determine the refractive index of benzene relative to water. 1.991.131.531.03 If the refractive index of diamond with respective air is 2.5, what will be the refractive index of air with respective diamond?2.50.50.40.8 Ultraviolet light with a wavelength of 300 nm passes from air into a substance with a refractive index of 1.5. What is the wavelength of light inside substance in AoAo ? (Given that 1𝑛𝑚=10Ao1nm=10 Ao )2300Ao2300 Ao 2100Ao2100 Ao 2000Ao2000 Ao None of the above Light travelling through a certain optical fiber at a speed of 1.95×108𝑚/𝑠1.95×10 8 m/s enters a different fiber where its speed changes to 2.30×108𝑚/𝑠2.30×10 8 m/s. If the refractive index of the first fiber is 1.48, determine the refractive index of the second fiber relative to the first. Correct up to 2 decimal points.1.180.850.120.02 Light travels through a 5 mm thick sheet of transparent plastic with a refractive index of 1.20. Calculate the time taken by the light to pass through the plastic sheet and emerge outside from other side. (Given: Speed of light in vacuum = 3×108𝑚/𝑠3×10 8 m/s)3×10−8𝑠3×10 −8 s2×10−8𝑠2×10 −8 s3×10−11𝑠3×10 −11 s2×10−11𝑠2×10 −11 s Light from a laser pointer strikes a plane mirror at an angle of 30o 30 o to the normal. If the angle of incidence decreases by 8o 8 o , what is the new angle of reflection?30o30 o 38o 38 o 22o 22 o 45o 45 o A 3.9 cm tall toy is placed at a distance of 24 cm from a concave mirror. If the image of toy formed is one-third the size of the toy and is inverted, determine the height of the image and focal length of the mirror.-1.3 cm and +6 cm respectively-1.3 cm and -6 cm respectively-1.3 cm and -12 cm respectively-1.3 cm and +12 cm respectively
Solution
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The refractive index of benzene relative to water can be calculated using the formula n21 = n2/n1. Here, n2 is the refractive index of benzene (1.50) and n1 is the refractive index of water (1.33). So, n21 = 1.50/1.33 = 1.13.
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The refractive index of air with respect to diamond is the reciprocal of the refractive index of diamond with respect to air. So, it is 1/2.5 = 0.4.
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The wavelength of light inside the substance can be calculated using the formula λ2 = λ1/n, where λ1 is the initial wavelength and n is the refractive index. Here, λ1 = 300 nm = 3000 Ao and n = 1.5. So, λ2 = 3000/1.5 = 2000 Ao.
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The refractive index of the second fiber relative to the first can be calculated using the formula n21 = v1/v2, where v1 and v2 are the speeds of light in the first and second fibers respectively. Here, v1 = 1.95×10^8 m/s and v2 = 2.30×10^8 m/s. So, n21 = 1.95/2.30 = 0.85 (rounded to two decimal points).
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The time taken by the light to pass through the plastic sheet can be calculated using the formula t = d/v, where d is the thickness of the sheet and v is the speed of light in the plastic. Here, d = 5 mm = 5×10^-3 m and v = c/n, where c is the speed of light in vacuum (3×10^8 m/s) and n is the refractive index of the plastic (1.20). So, v = 3×10^8/1.20 = 2.5×10^8 m/s and t = 5×10^-3/2.5×10^8 = 2×10^-11 s.
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The angle of reflection is equal to the angle of incidence. So, if the angle of incidence decreases by 8o, the new angle of reflection is also 30o - 8o = 22o.
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If the image of the toy is one-third the size of the toy and is inverted, the height of the image is -1.3 cm (negative because the image is inverted). The focal length of the mirror can be calculated using the mirror formula 1/f = 1/v - 1/u, where f is the focal length, v is the image distance and u is the object distance. Here, v = -3u (because the image is one-third the size of the object and is inverted) and u = -24 cm (negative because the object is on the same side of the mirror as the light source). So, 1/f = 1/(-3u) - 1/u = -4/(3u). Solving for f gives f = -3u/4 = -3(-24)/4 = +18 cm. However, because the image is inverted, the focal length is actually negative, so the correct answer is -1.3 cm and -18 cm respectively.
Similar Questions
A beam of light travels from air into a diamond at an angle of incidence 40o40 o . Inside the diamond, it bends and emerges at an angle of 25o25 o to the normal. Calculate the refractive index of the diamond.(Given: sin40o=0.643sin40 o =0.643 and sin25o=0.422sin25 o =0.422)1.451.521.621.72 A ray of light moves from a vacuum into a transparent substance. If the refractive index of the substance is 1.968 and the angle of incidence is 80o 80 o then find angle of refraction inside the substance. (Given: sin80o=0.984 sin80 o =0.984)60o60 o 45o 45 o 30o 30 o 90o 90 o The absolute refractive index of benzene is 1.50 and that of water is 1.33. Determine the refractive index of benzene relative to water. 1.991.131.531.03 If the refractive index of diamond with respective air is 2.5, what will be the refractive index of air with respective diamond?2.50.50.40.8 Ultraviolet light with a wavelength of 300 nm passes from air into a substance with a refractive index of 1.5. What is the wavelength of light inside substance in AoAo ? (Given that 1𝑛𝑚=10Ao1nm=10 Ao )2300Ao2300 Ao 2100Ao2100 Ao 2000Ao2000 Ao None of the above Light travelling through a certain optical fiber at a speed of 1.95×108𝑚/𝑠1.95×10 8 m/s enters a different fiber where its speed changes to 2.30×108𝑚/𝑠2.30×10 8 m/s. If the refractive index of the first fiber is 1.48, determine the refractive index of the second fiber relative to the first. Correct up to 2 decimal points.1.180.850.120.02 Light travels through a 5 mm thick sheet of transparent plastic with a refractive index of 1.20. Calculate the time taken by the light to pass through the plastic sheet and emerge outside from other side. (Given: Speed of light in vacuum = 3×108𝑚/𝑠3×10 8 m/s)3×10−8𝑠3×10 −8 s2×10−8𝑠2×10 −8 s3×10−11𝑠3×10 −11 s2×10−11𝑠2×10 −11 s Light from a laser pointer strikes a plane mirror at an angle of 30o 30 o to the normal. If the angle of incidence decreases by 8o 8 o , what is the new angle of reflection?30o30 o 38o 38 o 22o 22 o 45o 45 o A 3.9 cm tall toy is placed at a distance of 24 cm from a concave mirror. If the image of toy formed is one-third the size of the toy and is inverted, determine the height of the image and focal length of the mirror.-1.3 cm and +6 cm respectively-1.3 cm and -6 cm respectively-1.3 cm and -12 cm respectively-1.3 cm and +12 cm respectively A passenger in a bus looks at the rear view convex mirror with a focal length of 15 cm. If a toy is hanging at a distance of 30 cm from the mirror, where will the image of the toy appear and what will be the nature of the image?+10 cm; virtual, erect and diminished+15 cm; virtual, erect and enlarged-10 cm; real, erect and diminished-15 cm; virtual, erect and enlarged
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