Question 1:
easyThe frequency of a light wave in a material is $2 \times 10^{14} Hz$ and wavelength is $5000\mathring{A}$ . The refractive index of material will be:
(2007)
Velocity of light in the material is given by $v = f \lambda = (2 \times 10^{14} Hz) \times (5000 \times 10^{-10} m) = 10^8 m/s$ . The refractive index of the material is calculated as $n = \frac{c}{v} = \frac{3 \times 10^8}{10^8} = 3.00$ .
Question 2:
easyA ray of light travelling in air haves wavelength $\lambda$ , frequency n, velocity v and intensity, I. If this ray enters into water than these parameter are $\lambda’$ , n’, v’ and I’ respectively. Which relation is correct:
(2001)
When a light ray travels from one medium to another, its frequency remains absolutely unchanged because frequency is a fundamental characteristic of the source. Velocity and wavelength change based on the refractive index. Thus, $n = n'$ .
Question 3:
easyThe refractive index of water is 1.33. What will be the speed of light in water?
(1996)
The speed of light in a given medium is related to the refractive index by $v = \frac{c}{n}$ . Given the speed of light in vacuum $c = 3 \times 10^8 m/s$ and the refractive index of water $n = 1.33 = \frac{4}{3}$ . We find $v = \frac{3 \times 10^8}{4/3} = \frac{9 \times 10^8}{4} = 2.25 \times 10^8 m/s$ .
Question 4:
easyA light ray falls on a glass surface of refractive index $\sqrt{3}$, at an angle $60^\circ$. The angle between the refracted and reflected rays would be :
(2022)
By Snell's law, $1 \times \sin(60^\circ) = \sqrt{3} \times \sin(r)$, which gives $\sin(r) = 1/2$, so angle of refraction $r = 30^\circ$.
By law of reflection, angle of reflection is equal to angle of incidence $i = 60^\circ$.
Angle between reflected and refracted rays is $180^\circ - (i + r) = 180^\circ - (60^\circ + 30^\circ) = 90^\circ$.
Question 5:
easyA microscope is focused on a mark on a piece of paper and then a slab of glass of thickness $3 cm$ and refractive index $1.5$ is placed over the mark. How should the microscope be moved to get the mark in focus again?
(2006)
When a glass slab is placed over a mark, the mark appears raised by a distance known as normal shift.
Normal shift $\Delta t = t(1 - \frac{1}{\mu}) = 3(1 - \frac{1}{1.5}) = 3(1 - \frac{2}{3}) = 1 cm$.
To focus again on the apparent image, the microscope must be moved $1 cm$ upward.
Question 6:
moderateA beam of light composed of red and green rays is incident obliquely at a point on the face of a rectangular glass slab. When coming out on the opposite parallel face, the red and green rays emerge from:
(2004)
When composite light is incident obliquely on a parallel-sided glass slab, different colors undergo different refraction angles due to dispersion.
They suffer different lateral shifts but emerge parallel to the incident ray direction.
Thus, red and green rays emerge from two points propagating in two different parallel directions.
Question 7:
easyLight travels through a glass plate of thickness t and having a refractive index $\mu$. If c is the velocity of light in vacuum, the time taken by light to travel this thickness of glass is
(1996)
The velocity of light in the glass medium is $v = \frac{c}{\mu}$.
Time taken to travel thickness t is given by $T = \frac{distance}{velocity} = \frac{t}{v}$.
Substituting the velocity, $T = \frac{t}{c/\mu} = \frac{\mu t}{c}$.
Question 8:
easyTime taken by sunlight to pass through a window of thickness $4 mm$ whose refractive index is $\frac{3}{2}$ is
(1993)
Velocity of light in the window glass is $v = \frac{c}{\mu} = \frac{3 \times 10^8}{3/2} = 2 \times 10^8 m/s$.
Thickness is $t = 4 mm = 4 \times 10^{-3} m$.
Time taken is $T = \frac{t}{v} = \frac{4 \times 10^{-3}}{2 \times 10^8} = 2 \times 10^{-11} s$.
Question 9:
easyAn air bubble in a glass slab with refractive index $1.5$ (near normal incidence) is $5 cm$ deep when viewed from one surface and $3 cm$ deep when viewed from the opposite surface. The thickness (in cm) of the slab is:
(2016 – II)
Total apparent depth of the air bubble is the sum of apparent depths from both sides: $d_{app} = 5 + 3 = 8 cm$.
Apparent depth is related to real thickness T by $d_{app} = \frac{T}{\mu}$.
Therefore, real thickness $T = d_{app} \times \mu = 8 \times 1.5 = 12 cm$.
Question 10:
easyA bubble in glass slab ($\mu = 1.5$) when viewed from one side appears at $5 cm$ and $2 cm$ from other side, then thickness of slab is: (2000)
Total apparent thickness of the glass slab is $d_{app} = 5 + 2 = 7 cm$.
Using the relation $d_{app} = \frac{t}{\mu}$, the real thickness is $t = d_{app} \times \mu$.
Substituting the given values, $t = 7 \times 1.5 = 10.5 cm$.