NCERT Exemplar · Class 10 Science Light - Reflection and Refraction

Two rays fix each image: one parallel to the axis reflects through \(\displaystyle F\); one along a radius (through \(\displaystyle C\)) reflects back along itself. In (c) use the ray through \(\displaystyle F\) instead, which reflects parallel to the axis.(a) \(\displaystyle P\)–\(\displaystyle F\) \(\displaystyle \Rightarrow\) virtual, erect, magnified, behind the mirror. (b) \(\displaystyle F\)–\(\displaystyle C\) \(\displaystyle \Rightarrow\) real, inverted, magnified, beyond \(\displaystyle C\). (c) \(\displaystyle C\) \(\displaystyle \Rightarrow\) real, inverted, same size, at \(\displaystyle C\). (d) beyond \(\displaystyle C\) \(\displaystyle \Rightarrow\) real, inverted, diminished, between \(\displaystyle F\) and \(\displaystyle C\). (e) \(\displaystyle \infty\) \(\displaystyle \Rightarrow\) real, inverted, highly diminished, point at \(\displaystyle F\).Answer: inside \(\displaystyle F\): virtual, magnified, behind the mirror; \(\displaystyle F\) to \(\displaystyle C\): magnified, beyond \(\displaystyle C\); at \(\displaystyle C\): same size; beyond \(\displaystyle C\): diminished, between \(\displaystyle F\) and \(\displaystyle C\); at infinity: point image at \(\displaystyle F\).

Two rays fix each image: one parallel to the axis refracts through \(\displaystyle F_2\); one through the optical centre \(\displaystyle O\) passes straight through undeviated.(a) Between \(\displaystyle O\) and \(\displaystyle F_1\): virtual, erect, magnified, same side as object. (b) Between \(\displaystyle F_1\) and \(\displaystyle 2F_1\): real, inverted, magnified, beyond \(\displaystyle 2F_2\). (c) At \(\displaystyle 2F_1\): real, inverted, same size, at \(\displaystyle 2F_2\). (d) At infinity: real, inverted, highly diminished, at \(\displaystyle F_2\). (e) At \(\displaystyle F_1\): real, inverted, highly magnified, at infinity.Answer: inside \(\displaystyle F_1\): virtual, erect, magnified; \(\displaystyle F_1\) to \(\displaystyle 2F_1\): magnified, beyond \(\displaystyle 2F_2\); at \(\displaystyle 2F_1\): same size, at \(\displaystyle 2F_2\); at infinity: point image at \(\displaystyle F_2\); at \(\displaystyle F_1\): image at infinity.

Laws of refraction: 1. The incident ray, the refracted ray and the normal at the point of incidence lie in one plane. 2. For a given pair of media and a given colour of light, \[\frac{\sin i}{\sin r} = n \] a constant (Snell's law), \(\displaystyle i\) and \(\displaystyle r\) being the angles of incidence and refraction.The slab's faces are parallel, so the angle of refraction at \(\displaystyle A\) is the angle of incidence at \(\displaystyle B\): \[\text{at } A \text{ (air}\to\text{glass)}:\quad \frac{\sin i}{\sin r} = n \] \[\text{at } B \text{ (glass}\to\text{air)}:\quad \frac{\sin r}{\sin e} = \frac{1}{n} \] \[\Rightarrow\ \sin e = n\sin r = \sin i \ \Rightarrow\ e = i \] So the emergent ray is parallel to the incident ray, displaced sideways.Answer: the incident ray, refracted ray and normal are coplanar and \(\displaystyle \sin i/\sin r = n\); through a glass slab \(\displaystyle e = i\), so the emergent ray is parallel to the incident ray, laterally displaced.

A concave lens always forms a virtual, erect, diminished image between \(\displaystyle O\) and \(\displaystyle F_1\) on the object's side, whatever the object distance. Rays: one parallel to the axis refracts as if diverging from \(\displaystyle F_1\); one through \(\displaystyle O\) passes straight.(a) At \(\displaystyle F_1\): image between \(\displaystyle O\) and \(\displaystyle F_1\), virtual, erect, diminished. (b) Between \(\displaystyle F_1\) and \(\displaystyle 2F_1\): image nearer \(\displaystyle F_1\) than in (a), virtual, erect, diminished. (c) Beyond \(\displaystyle 2F_1\): image nearer \(\displaystyle F_1\) still, virtual, erect, smaller again.Answer: in all three cases virtual, erect, diminished, between \(\displaystyle O\) and \(\displaystyle F_1\) on the object's side, moving towards \(\displaystyle F_1\) and shrinking as the object recedes.

A convex mirror always forms a virtual, erect, diminished image between \(\displaystyle P\) and \(\displaystyle F\), whatever the object distance. Rays: one parallel to the axis reflects as if diverging from \(\displaystyle F\) behind the mirror; one aimed at \(\displaystyle C\) behind the mirror reflects back along itself.(a) At infinity: image at \(\displaystyle F\), virtual, erect, point-sized. (b) At a finite distance: image between \(\displaystyle P\) and \(\displaystyle F\), virtual, erect, diminished, larger than in (a) as the object nears the mirror.Answer: (a) point-sized, virtual, erect image at \(\displaystyle F\) behind the mirror; (b) virtual, erect, diminished image between \(\displaystyle P\) and \(\displaystyle F\), behind the mirror.

Magnification for a real, inverted image: \(\displaystyle m = \dfrac{v}{u} = -3\), with \(\displaystyle v = +80\ \mathrm{cm}\). \[u = \dfrac{v}{m} = \dfrac{80}{-3} = -\dfrac{80}{3}\ \mathrm{cm} \approx -26.7\ \mathrm{cm} \] Lens formula: \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{80} - \dfrac{1}{-80/3} = \dfrac{1}{80} + \dfrac{3}{80} = \dfrac{4}{80}\ \mathrm{cm^{-1}} \] \[f = +20\ \mathrm{cm} \] \(\displaystyle f\) positive \(\displaystyle \Rightarrow\) convex lens.Answer: candle \(\displaystyle \tfrac{80}{3}\approx 26.7\ \mathrm{cm}\) from the lens; image real, inverted, magnified $\displaystyle 3$ times; lens convex.

Mirror formula \(\displaystyle \dfrac1v+\dfrac1u=\dfrac1f\), magnification \(\displaystyle m=-\dfrac{v}{u}\). A diminished image has two possible mirrors.Convex mirror, \(\displaystyle f=+20\ \mathrm{cm}\), image virtual and erect \(\displaystyle \Rightarrow m=+\dfrac13\): \[-\dfrac{v}{u}=\dfrac13 \Rightarrow v=-\dfrac{u}{3} \] \[\dfrac{1}{-u/3}+\dfrac1u=-\dfrac{2}{u}=\dfrac{1}{20}\ \mathrm{cm^{-1}} \Rightarrow u=-40\ \mathrm{cm} \]Concave mirror, \(\displaystyle f=-20\ \mathrm{cm}\), image real and inverted \(\displaystyle \Rightarrow m=-\dfrac13\): \[-\dfrac{v}{u}=-\dfrac13 \Rightarrow v=\dfrac{u}{3} \] \[\dfrac{3}{u}+\dfrac1u=\dfrac{4}{u}=-\dfrac{1}{20}\ \mathrm{cm^{-1}} \Rightarrow u=-80\ \mathrm{cm} \] \[|u|=80\ \mathrm{cm}>2|f|=40\ \mathrm{cm} \] Object beyond C, consistent with a diminished real image.Answer: concave mirror, object \(\displaystyle 80\ \mathrm{cm}\) from the mirror, image real, inverted, diminished (a convex mirror would need the object at \(\displaystyle 40\ \mathrm{cm}\), image virtual and erect).

Power of a lens is the reciprocal of its focal length in metres: \[P = \dfrac{1}{f\,(\mathrm{m})} \] Unit: dioptre (D), \(\displaystyle 1\ \mathrm{D} = 1\ \mathrm{m^{-1}}\).First student, \(\displaystyle f=+50\ \mathrm{cm}=+0.5\ \mathrm{m}\) \(\displaystyle \Rightarrow\) convex (converging) lens: \[P_1=\dfrac{1}{0.5\ \mathrm{m}}=+2\ \mathrm{D} \]Second student, \(\displaystyle f=-50\ \mathrm{cm}=-0.5\ \mathrm{m}\) \(\displaystyle \Rightarrow\) concave (diverging) lens: \[P_2=\dfrac{1}{-0.5\ \mathrm{m}}=-2\ \mathrm{D} \]Answer: \(\displaystyle P = 1/f\) (\(\displaystyle f\) in m), unit dioptre; first student: convex lens, \(\displaystyle +2\ \mathrm{D}\); second student: concave lens, \(\displaystyle -2\ \mathrm{D}\).