Refraction of Light at Plane Surfaces — ICSE Class 10 Physics Important Questions

Answer: (b) both decrease.

Glass is optically denser, so light slows down. Since frequency is unchanged, $v=f\lambda$v=f means the wavelength also decreases in glass.

Answer: (c) no unit.

Refractive index is a ratio of two speeds (or two like quantities), so the units cancel and it is a pure number.

Answer: (b) $\sin C=\dfrac{3}{4}$C=3/4.

For a denser-to-rarer interface, $\sin C=\dfrac{1}{n}=\dfrac{1}{4/3}=\dfrac{3}{4}$C=1/n=1/4/3=3/4, giving $C\approx48.6^\circ$C48.6^.

Answer: (a) $12\text{ cm}$12 cm.

$n=\dfrac{\text{real depth}}{\text{apparent depth}}\Rightarrow \text{apparent depth}=\dfrac{16}{4/3}=16\times\dfrac{3}{4}=12\text{ cm}.$n=real depth/apparent depth apparent depth=16/4/3=16×3/4=12 cm. The coin appears raised by $4\text{ cm}$4 cm.

Answer: (a) Diamond's high refractive index ($n\approx2.42$n2.42) gives a small critical angle ($\approx24^\circ$24^), so light entering it is repeatedly totally internally reflected before emerging, making it sparkle. R correctly explains A.

First law: The incident ray, the refracted ray and the normal at the point of incidence all lie in the same plane.

Second law (Snell's law): For two given media and light of a given colour, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant, equal to the refractive index of the second medium with respect to the first:
$\frac{\sin i}{\sin r}=n\ (\text{constant}).$i/ r=n (constant).

Refractive index $n=\dfrac{c}{v}$n=c/v, so
$v=\frac{c}{n}=\frac{3\times10^{8}}{1.5}=2\times10^{8}\text{ m s}^{-1}.$v=c/n=3×10^81.5=2×10^8 m s^-1.

The speed of light in glass is $2\times10^{8}\text{ m s}^{-1}$2×10^8 m s^-1.

By Snell's law for air to glass:
$n=\frac{\sin i}{\sin r}\Rightarrow \sin r=\frac{\sin i}{n}.$n= i/ r r= i/n.

$\sin r=\frac{\sin60^\circ}{1.5}=\frac{0.866}{1.5}=0.577.$r=60^/1.5=0.866/1.5=0.577.

$r=\sin^{-1}(0.577)\approx35.3^\circ.$r=^-1(0.577)35.3^.

The angle of refraction is about $35^\circ$35^, the ray bending towards the normal as it enters the denser glass.

The ray bends towards the normal on entering the slab and away from the normal on leaving; the emergent ray is parallel to the incident ray but shifted sideways by the lateral displacement $d$d.

The lateral displacement (perpendicular distance between the incident ray produced and the emergent ray) depends on: (i) the thickness of the slab (greater thickness gives greater shift), and (ii) the angle of incidence (larger angle gives greater shift); it also increases with the refractive index of the glass.

(i) For the denser-to-rarer (medium-to-air) interface:
$\sin C=\frac{1}{n}\Rightarrow n=\frac{1}{\sin C}=\frac{1}{\sin30^\circ}=\frac{1}{0.5}=2.$C=1/n n=1/ C=1/30^=1/0.5=2.

(ii) Speed of light in the medium:
$v=\frac{c}{n}=\frac{3\times10^{8}}{2}=1.5\times10^{8}\text{ m s}^{-1}.$v=c/n=3×10^82=1.5×10^8 m s^-1.

The refractive index is $2$2 and the speed of light in the medium is $1.5\times10^{8}\text{ m s}^{-1}$1.5×10^8 m s^-1.

(a) Total internal reflection is the complete reflection of a light ray back into a denser medium when it strikes the boundary with a rarer medium at an angle greater than the critical angle. Conditions: (i) light must travel from a denser to a rarer medium; (ii) the angle of incidence must exceed the critical angle $C$C.

(b) A right-angled isosceles prism turns a ray through $90^\circ$90^ by total internal reflection at its hypotenuse.

(c) Critical angle $C\approx41.8^\circ$C41.8^. Since the angle of incidence inside the glass is $45^\circ$45^, and $45^\circ>41.8^\circ$45^>41.8^, the light strikes at more than the critical angle, so total internal reflection does occur. This is why such prisms are used in periscopes and binoculars.

(a) Speed $v=\dfrac{c}{n}$v=c/n, so light is faster in the medium with the smaller refractive index. Since $n_w=1.33<n_g=1.5$n_w=1.33<n_g=1.5, light travels faster in water.

(b) Refractive index of glass with respect to water:
$_w n_g=\frac{n_g}{n_w}=\frac{1.5}{1.33}=1.128.$_w n_g=n_g/n_w=1.5/1.33=1.128.

(c) Applying Snell's law at the water-glass boundary:
$n_w\sin i=n_g\sin r$n_w i=n_g r
$1.33\times\sin30^\circ=1.5\times\sin r$1.33×30^=1.5× r
$1.33\times0.5=1.5\times\sin r\Rightarrow \sin r=\frac{0.665}{1.5}=0.443.$1.33×0.5=1.5× r r=0.665/1.5=0.443.
$r=\sin^{-1}(0.443)\approx26.3^\circ.$r=^-1(0.443)26.3^.

The ray bends towards the normal (from $30^\circ$30^ to about $26^\circ$26^) on entering the denser glass.

(i) Both observations are caused by the refraction of light at the water-air surface.

(ii) Light travels from water (denser) to air (rarer), so it bends away from the normal as it leaves the water.

(iii) Using $n=\dfrac{\text{real depth}}{\text{apparent depth}}$n=real depth/apparent depth:
$\text{apparent depth}=\frac{\text{real depth}}{n}=\frac{20}{4/3}=20\times\frac{3}{4}=15\text{ cm}.$apparent depth=real depth/n=20/4/3=20×3/4=15 cm.

(iv) The coin appears raised by
$20-15=5\text{ cm}.$20-15=5 cm.