Question
From the given data set, determine angular dispersion by the prism and dispersive power of its material for extreme colours. $nR = 1.62 nv = 1.66, δR = 3.1^\circ $
✓
Answer
Given: $n_R = 1.62, n_V = 1.66, δ_R = 3.1^\circ $
To find:
i. Angular dispersion $(δ_{vr})$
ii. Dispersive power $(ω_{VR})$
Formula:
i. $δ = A (n – 1)$
ii.$δ_{VR} = δ_V – δ_R$
(iii) $\omega=\frac{\delta_{ V }-\delta_{ R }}{\left(\frac{\delta_{ V }+\delta_{ R }}{2}\right)}$
Calculation: From formula (i),
$ \delta_{ R }= A \left( n _{ R }-1\right)$
$\therefore A =\frac{\delta_R}{\left(n_R-1\right)}=\frac{3.1}{(1.62-1)}=\frac{3.1}{0.62}$
$=5$
$\delta V = A \left( n _{ V }-1\right)=5 \times(1.66-1)=3.3 C $
From formula (ii),
$\delta_{ VR }=3.3-3.1=0.2^{\circ}$
From formula (iii),
$ \omega_{V R}=\frac{3.3-3.1}{\left(\frac{3.3+3.1}{2}\right)}=\frac{0.2}{6.4} \times 2=\frac{0.2}{3.2}=\frac{1}{16}$
$=0.0625 $
To find:
i. Angular dispersion $(δ_{vr})$
ii. Dispersive power $(ω_{VR})$
Formula:
i. $δ = A (n – 1)$
ii.$δ_{VR} = δ_V – δ_R$
(iii) $\omega=\frac{\delta_{ V }-\delta_{ R }}{\left(\frac{\delta_{ V }+\delta_{ R }}{2}\right)}$
Calculation: From formula (i),
$ \delta_{ R }= A \left( n _{ R }-1\right)$
$\therefore A =\frac{\delta_R}{\left(n_R-1\right)}=\frac{3.1}{(1.62-1)}=\frac{3.1}{0.62}$
$=5$
$\delta V = A \left( n _{ V }-1\right)=5 \times(1.66-1)=3.3 C $
From formula (ii),
$\delta_{ VR }=3.3-3.1=0.2^{\circ}$
From formula (iii),
$ \omega_{V R}=\frac{3.3-3.1}{\left(\frac{3.3+3.1}{2}\right)}=\frac{0.2}{6.4} \times 2=\frac{0.2}{3.2}=\frac{1}{16}$
$=0.0625 $
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find the acceleration due to gravity on a planet that is $10$ times as massive as the Earth and with radius $20$ times of the radius of the Earth $(g = 9.8 \ m/s^2).$
→What happens if a bar magnet is cut into two pieces transverse to its length/ along its length?
What is the importance of energy gap in a semiconductor?
Calculate the value of the universal gravitational constant from the given data. Mass of the Earth $= 6 \times 10^{24}\ kg$, Radius of the Earth $= 6400\ km$ and the acceleration due to gravity on the surface $= 9.8 m/s^2$
→If the height of a $TV$ transmitting antenna is $128 m,$ how much square area can be covered by the transmitted signal if the receiving antenna is at the ground level? $($Radius of the Earth $= 6400 \ km)$
→A gas at $900 ^{}\circ C$ is cooled until both its pressure and volume are halved. Calculate its final temperature.
→Mention some main characteristics of gravitational force.
If $g =9.8 m / s ^2$ on the surface of the Earth, find its value at $h =\frac{ R }{2}$ from the surface of the Earth.
→Calculate
$i.$ total resistance and
$ii.$ total current in the following circuit.
$R _1=3 \Omega, R _2=6 \Omega, R _3=5 \Omega, V =14 V$
→$i.$ total resistance and
$ii.$ total current in the following circuit.
$R _1=3 \Omega, R _2=6 \Omega, R _3=5 \Omega, V =14 V$
If $|\overrightarrow{ A }+\overrightarrow{ B }|=|\overrightarrow{ A }-\overrightarrow{ B }|$ then what can be the angle between $\overrightarrow{ A }$ and $\overrightarrow{ B }$ ?
→