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CalibratePhysics

9.3

Refraction

Ray Optics and Optical Instruments · CBSE

Theory

What this subtopic is examined on

  • Statement of the laws of refraction and the meaning of refractive index.
  • Relations between refractive index, speed and wavelength, and the fact that frequency is unchanged.
  • Relative refractive index, its reciprocal relation and the chain rule.
  • The parallel-sided slab: no deviation, only lateral shift.
  • Apparent depth and the normal shift produced by a slab.
  • Numericals converting between nn, cc, vv, λ\lambda and ν\nu.

The phenomenon

  • When light meets a boundary with another transparent medium, part is reflected back and part enters the second medium. An obliquely incident ray (0∘<i<90∘0^\circ < i < 90^\circ) changes its direction of propagation at the interface: this is refraction.
  • The angles of incidence and refraction are always measured from the normal to the interface at the point of incidence.

Laws of refraction (Snell's laws)

  • First law: the incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the same plane.
  • Second law: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media:

sin⁡isin⁡r=n21\dfrac{\sin i}{\sin r} = n_{21}

  • n21n_{21} is the refractive index of medium 2 with respect to medium 1. It is a characteristic of the pair of media and depends on the wavelength of the light, but it is independent of the angle of incidence.
  • The symmetric form, needed whenever neither medium is air:

n1sin⁡i=n2sin⁡rn_1\sin i = n_2\sin r

Refractive index, speed and wavelength

n21=n2n1=v1v2=λ1λ2,n=cvn_{21} = \dfrac{n_2}{n_1} = \dfrac{v_1}{v_2} = \dfrac{\lambda_1}{\lambda_2}, \qquad n = \dfrac{c}{v}

  • nn is the absolute refractive index of a medium: the ratio of the speed of light in vacuum to that in the medium. It is always greater than 1 for a material medium.
  • On passing into a denser medium the speed falls and the wavelength shortens in the same ratio, but the frequency is fixed by the source and does not change:

ν=cλvacuum=vλmedium\nu = \dfrac{c}{\lambda_{vacuum}} = \dfrac{v}{\lambda_{medium}}

  • Refractive index also depends on wavelength: it is larger for violet than for red (normal dispersion), which is why a prism disperses white light.

Optically denser and rarer

  • If n21>1n_{21} > 1 then r<ir < i: the ray bends towards the normal and medium 2 is optically denser than medium 1.
  • If n21<1n_{21} < 1 then r>ir > i: the ray bends away from the normal; this is light going from a denser medium into a rarer one.
  • Optical density is not mass density. Turpentine is optically denser than water although its mass density is lower. Optical density compares speeds of light, not masses per unit volume.

Relations between relative indices

n12=1n21n32=n31×n12n_{12} = \dfrac{1}{n_{21}} \qquad\qquad n_{32} = n_{31}\times n_{12}

  • The second is the chain rule that lets a question about a water–glass interface be answered from separate air–glass and air–water data:

nglass, water=nglassnwatern_{glass,\,water} = \dfrac{n_{glass}}{n_{water}}

The parallel-sided glass slab

Diagram in words

  • A rectangular glass slab of thickness tt with a ray incident on the top face at angle i1i_1, refracting into the slab at r1r_1, travelling to the lower face where it strikes at i2=r1i_2 = r_1, and emerging at r2r_2. Draw the dashed extension of the original incident ray to show the sideways displacement of the emergent ray from it.

Results

  • Refraction happens twice, air–glass and glass–air, and the two refractions are exactly opposite, so r2=i1r_2 = i_1: the emergent ray is parallel to the incident ray.
  • There is therefore no net deviation, only a lateral (sideways) displacement:

d=t sin⁡(i−r)cos⁡rd = \dfrac{t\,\sin(i-r)}{\cos r}

  • The shift grows with the thickness of the slab, with the angle of incidence, and with the refractive index. At normal incidence it is zero.

Apparent depth

  • For viewing close to the normal, an object at real depth h2h_2 under a medium of refractive index nn appears to be at depth

h1=h2n⇒n=real depthapparent depthh_1 = \dfrac{h_2}{n} \qquad\Rightarrow\qquad n = \dfrac{\text{real depth}}{\text{apparent depth}}

  • The bottom of a water tank therefore appears raised. Looking obliquely, the object also appears laterally shifted.
  • A slab of thickness tt and index nn laid over an object raises it by

normal shift=t(1−1n)\text{normal shift} = t\left(1 - \dfrac{1}{n}\right)

  • The shift does not depend on where between the object and the eye the slab is held.

Worked pattern CBSE has set

  • Monochromatic light of λair=589\lambda_{air} = 589 nm falls from air on water of μ=1.33\mu = 1.33. Frequency is unchanged:

ν=cλair=3×1085.89×10−7=5.09×1014 Hz\nu = \dfrac{c}{\lambda_{air}} = \dfrac{3\times10^{8}}{5.89\times10^{-7}} = 5.09\times10^{14}\ \text{Hz}

v=cμ=3×1081.33≈2.26×108 m s−1v = \dfrac{c}{\mu} = \dfrac{3\times10^{8}}{1.33} \approx 2.26\times10^{8}\ \text{m s}^{-1}

λwater=λairμ=5.89×10−71.33≈4.43×10−7 m\lambda_{water} = \dfrac{\lambda_{air}}{\mu} = \dfrac{5.89\times10^{-7}}{1.33} \approx 4.43\times10^{-7}\ \text{m}

  • The reflected light, in contrast, keeps the wavelength, frequency and speed it had in air.

Standard results

  • sin⁡i/sin⁡r\sin i/\sin r is constant for a given pair of media and a given colour; it is not constant if the colour changes.
  • n=c/vn = c/v, so a medium with v=2×108v = 2\times10^{8} m s−1^{-1} has n=1.5n = 1.5.
  • Frequency is the invariant on refraction; wavelength and speed are not.
  • A slab shifts but never deviates; a prism deviates.
  • Real depth exceeds apparent depth by the factor nn.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

  1. 1

    Monochromatic light of wavelength 589 nm is incident from air on a water surface. If μ\mu for water is 1.33, find the wavelength, frequency and speed of the refracted light.

    3 marksCBSE Class 12 Physics — SA · 2/3, AI 2017, Ap · Section C, 3 marksAnswer 1 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

  1. Given: wavelength in air λa=589\lambda_a=589 nm =5.89×10−7=5.89\times10^{-7} m, refractive index of water μw=1.33\mu_w=1.33, speed of light in vacuum c=3×108c=3\times10^8 m/s. Frequency (unchanged on refraction):   ν=cλa=3×1085.89×10−7=5.093×1014 Hz\;\nu=\dfrac{c}{\lambda_a}=\dfrac{3\times10^8}{5.89\times10^{-7}}=5.093\times10^{14}\text{ Hz} Speed of light in water:   v=cμw=3×1081.33≈2.26×108 m/s\;v=\dfrac{c}{\mu_w}=\dfrac{3\times10^8}{1.33}\approx2.26\times10^8\text{ m/s} Wavelength in water:   λw=vν=λaμw=5.89×10−71.33≈4.43×10−7 m\;\lambda_w=\dfrac{v}{\nu}=\dfrac{\lambda_a}{\mu_w}=\dfrac{5.89\times10^{-7}}{1.33}\approx4.43\times10^{-7}\text{ m} Thus, for the refracted light: wavelength ≈4.43×10−7\approx4.43\times10^{-7} m, frequency =5.09×1014=5.09\times10^{14} Hz, and speed ≈2.26×108\approx2.26\times10^8 m/s.