Makalu

Snell's law: Lissafin refraction

  • Makala ta da ta gabata wadda ke dauke da tarihin snell’s law wadda na yi bayanin cewa Ibn sahl shine mutum na farko da ya kawo wannan Snell's law kamar yadda muka san shi a yau. A bayanin kuma na kawo fomuloli wanda ake amfani da su wajen lissafin refraction. A wannan karon, yanzu zamu dauki misalai daga cikin tambayoyin jarabawar karshen aji a sakandire wato JAMB da WAEC.

    Tambaya ta farko:

    1. A beam of light is incident from air to water "t" an angle of 300. Find the angle of refraction if the refractive index of water is 4/3.  A.50    B.18  C.220      D.240  .      (JAMB 1993)

    Amsar tambaya:

    Zamu fara fitar da abubuwan da aka bamu kamun musan wacce fomula zamu yi amfani da ita

    Data:   

    Angle of incidence,  i = 300

    Refractive index, ∩ = 4/3

    Angle of refraction, r = ?

    Yanzu zamu yi amfani da law na refraction, ga ta kamar haka: ∩ = sin I / sin r

    Data da muka fitar yanzu zamu sa kowannensu cikin fomula kamar haka:  4/3 = sin30 /sin r

    Yanzu zamu yi cross multiply saboda 'r' ta zama subject na fomula

    Sin r × 4 = 3 × sin 300

    Sin r = 3 × 0.5 / 4 = 1.5 / 4 = 0.375

    Yanzu angle of refraction zai zama:  r = sin-1 0.375 = 220

    Saboda haka amsarmu ya zama 'C' ke nan.

    Za a iya karanat: Abubuwan da ya kamata a sani game da mechanical energy da machin

    Tambaya ta biyu:

    1. Light of velocity 3.0 × 108ms-1 is incident on a material of refractive index ∩ . if the velocity of light is reduced to 2.4 × 108ms-1   in the material, what is ∩?    A. 3.33     B.  2.25    C.1.33   D.1.25   (JAMB 1997)   

    Amsar tambaya: 

    Bari mu fara fitarda abinda aka bamu

    Data:

    Velocity of light at incidence Va = 3.0 × 108ms-1

    Velocity of light after refraction Vg  = 2.4 ×  108ms-1  

    Yanzu zamu dauko fomula da zamu yi amfani da ita.

    Refractive index ∩ = Va / Vg = 3.0 × 108 / 2.4 ×  108  = 1.25

    Saboda haka amsarmu shine 'D' ke nan.

    Tambaya ta Uku:

    1. The velocities of light in air and glass are 3.0 × 108ms-1 and 1.8 × 108ms-1. Calculate the sine of the angle of incidence that will produce an angle of refraction of 300 for a ray of light incident on glass.     (WAEC 1988)

    Amsar tambaya:

    Yanzu zamu fara fitar da abubuwanda aka bamu

    Data:

    Velocity of light in Air,  Va = 3.0 × 108ms-1

    Velocity of light in glass, Vg = 1.8 × 108ms-1

    Angle of refraction  r = 300

    Sine of angle incidence, sin I = ?

    Yanzu zamu dauko formula da zamu yi amfani da ita wajan lissafin

    ag  = velocity of light in Air / velocity of light in Glass

    = sin I / sin 300  = 3.0 × 108  / 1.8 × 108

    Zai zama sin i =  sin 300  × 3.0 × 108  /  1.8 × 108  = 0.5 × 3 / 1.8 = 1.5 / 1.8

    = 0.83

    Tambaya ta hudu:

    1. An electromagnetic wave of frequency 5.0 × 1014 HZ is incident on the surface of water of refractive index 4/3. Taking the speed of wave in air as 3.0 × 108 ms-1 , calculate the wavelength of the wave in water. (WAEC 1998)

    Amsar tambaya:

    Data:

    Frequency, f = 5.0 × 1014 HZ

    Refractive index, ∩ = 4/3

    Wave speed, V = 3.0 × 108 ms-1  

    Daga V = λf, wavelength in air,  λ = V / f = 3.0 × 108  /  5.0 × 1014  = 6.0 × 10-7m

    Yanzu zamu dauko formula da zamu yi amfani da ita wajan lissafin kamar haka:

    aw = wavelength of wave in air (λa) / wavelength of wave in water (λw)

    4/3 = 6.0 × 10-7 / λw

    λw =  6.0 × 10-7 / 4 = 3 × 6.0 × 10-7 / 4

    = 4.50 × 10-7 m

    Wadannan tambayoyi misalai kadanne da na kawo muku. Yana da kyau dalibi ya kara karatu  da kuma gwada amsa wasu tambayoyin domin tabbatar da ya fahimci wannan karatun.

    Sannan mai karatu zai iya duba: Linear momentum: Misalin lissafin impulse

    Hakkin mallakar hoto:Byju's

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