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Light: Reflection and Refraction - REFRACTION OF LIGHT9.3 REFRACTION OF LIGHT9.3 REFRACTION OF LIGHT9.3 REFRACTION OF LIGHT9.3 REFRACTION OF LIGHT

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Refraction is the phenomenon of the bending of light when it passes obliquely from one transparent medium to another due to a change in the speed of light.

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When light travels from a rarer medium to a denser medium, it bends towards the normal. Conversely, it bends away from the normal when traveling from a denser to a rarer medium.

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The first law of refraction states that the incident ray, the refracted ray, and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.

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Snell's Law: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for light of a given color and for a given pair of media, expressed as sin⁡isin⁡r=n21\frac{\sin i}{\sin r} = n_{21}.

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The absolute refractive index nmn_m of a medium is given by the ratio of the speed of light in vacuum cc to the speed of light in the medium vv: nm=cvn_m = \frac{c}{v}.

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For a rectangular glass slab, the emergent ray is parallel to the incident ray but is displaced laterally. The angle of incidence ii is equal to the angle of emergence ee.

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Lens Formula: It gives the relationship between object distance uu, image distance vv, and focal length ff as 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}.

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Magnification mm produced by a lens is the ratio of the height of the image h′h' to the height of the object hh, and is also related to distances as m=h′h=vum = \frac{h'}{h} = \frac{v}{u}.

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Power of a lens PP is the reciprocal of its focal length ff in meters (P=1fP = \frac{1}{f}). The SI unit of power is Dioptre (DD).

📐Formulae

n21=sin⁡isin⁡rn_{21} = \frac{\sin i}{\sin r}

nm=cvn_m = \frac{c}{v}

n21=v1v2n_{21} = \frac{v_1}{v_2}

1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

m=h′h=vum = \frac{h'}{h} = \frac{v}{u}

P=1f(in meters)P = \frac{1}{f (\text{in meters})}

💡Examples

Problem 1:

The refractive index of glass is 1.501.50. If the speed of light in vacuum is 3×108 m/s3 \times 10^8 \text{ m/s}, calculate the speed of light in glass.

Solution:

Given: ng=1.50n_g = 1.50 and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}. Using the formula ng=cvgn_g = \frac{c}{v_g}, we get vg=cngv_g = \frac{c}{n_g}. Thus, vg=3×1081.50=2×108 m/sv_g = \frac{3 \times 10^8}{1.50} = 2 \times 10^8 \text{ m/s}.

Explanation:

The speed of light decreases when entering a denser medium. Here, the speed in glass is calculated by dividing the speed in vacuum by the absolute refractive index of glass.

Problem 2:

A concave lens has a focal length of 15 cm15 \text{ cm}. At what distance should the object from the lens be placed so that it forms an image at 10 cm10 \text{ cm} from the lens?

Solution:

For a concave lens, f=−15 cmf = -15 \text{ cm} and v=−10 cmv = -10 \text{ cm} (since image is virtual). Using 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f}: 1u=1v−1f\frac{1}{u} = \frac{1}{v} - \frac{1}{f} 1u=1−10−1−15\frac{1}{u} = \frac{1}{-10} - \frac{1}{-15} 1u=−110+115\frac{1}{u} = -\frac{1}{10} + \frac{1}{15} 1u=−3+230=−130\frac{1}{u} = \frac{-3 + 2}{30} = -\frac{1}{30} u=−30 cmu = -30 \text{ cm}

Explanation:

Applying the lens formula with the correct sign convention (ff and vv are negative for concave lenses forming virtual images), the object distance is found to be 30 cm30 \text{ cm} in front of the lens.

Problem 3:

Calculate the power of a convex lens of focal length 40 cm40 \text{ cm}.

Solution:

Given f=+40 cmf = +40 \text{ cm}. First, convert focal length to meters: f=40100=0.4 mf = \frac{40}{100} = 0.4 \text{ m} Now, use P=1fP = \frac{1}{f}: P=10.4=+2.5 DP = \frac{1}{0.4} = +2.5 \text{ D}

Explanation:

The power of a lens is defined as the reciprocal of its focal length in meters. Since the lens is convex, the focal length and power are positive.