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The angle of minimum deviation produced by a thin glass prism in air is 'δ'. If that prism is immersed in water, the angle of minimum deviation will be -

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प्रश्न

The angle of minimum deviation produced by a thin glass prism in air is 'δ'. If that prism is immersed in water, the angle of minimum deviation will be ______.

( aμg = refractive index of glass w.r.t, air aμw = refractive index of water w.r.t. air)

पर्याय

  • `delta[(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g+1))]`

  • `delta[(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g-1))]`

  • `delta[(""_amu_w-""_amu_g)/(""_amu_w(""_amu_g-1))]`

  • `delta[((""_amu_w-""_amu_g))/(""_amu_wxx""_amu_g)]`

MCQ
रिकाम्या जागा भरा

उत्तर

The angle of minimum deviation produced by a thin glass prism in air is 'δ'. If that prism is immersed in water, the angle of minimum deviation will be `underline(delta[(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g-1))])`

Explanation:

δ = (µ - 1)

∴ δair = aμg - 1

δwater = wμg - 1

`thereforedelta_"water"/delta_"air"=(""_wmu_g-1)/(""_amu_g-1)=((""_amu_g)/(""_amu_w)-1)/(""_amu_g-1)=(""_amu_g-""_amu_w)/(""_amu_w(""_amu_g-1))`

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