Numerical Problems

Question 15 Marks

A microscope has an objective and eyepiece of focal lengths $5 cm$ and $50 cm$ respectively with tube length $30 cm$. Find the magnification of the microscope in the near point and normal focusing.

Answer

$f _0=5 cm =5 \times 10^{-2} m ;$
$f _{ e }=50 cm =50 \times 10^{-2} m ;$
$L =30 cm =30 \times 10^{-2} m ;$
$D =25 cm =25 \times 10^{-2} m$
(i) The total magnification $m$ in near point focusing is,$m = m _0 m _{ e }=\left(\frac{L}{f_0}\right)\left(1+\frac{D}{f_0}\right)$
Substituting,
$m = m _0 m _{ e }=\left(\frac{30 \times 10^{-2}}{5 \times 10^{-3}}\right)\left(1+\frac{25 \times 10^{-2}}{50 \times 10^{-2}}\right)$
$=(6)(1.5)=9$
(ii) The total magnification $m$ in normal focusing is,$m = m _0 m _{ e }=\left(\frac{L}{f_0}\right)\left(\frac{D}{f_e}\right)$Substituting,

View full question & answer→

Question 25 Marks

The refractive index of a prism material is $1.541.$ Find its critical angle.

Answer

Given data:
$n =1.541 $
$i _{ C }=\sin ^{-1}\left(\frac{1}{n}\right) $
$=\sin ^{-1}\left(\frac{1}{1.541}\right) $
$i _{ c }=\sin ^{-1}(0.6489) $
$i _{ c }=40^{\circ} 5^{\prime}$

View full question & answer→

Question 35 Marks

An object is placed at a distance of $20.0 cm$ from a concave mirror of focal length $15.00 m$. What distance from the mirror a screens should be placed to get a sharp image? What is the nature of image? $F=-150 m$, $u$ $=-20 cm$

Answer

$F =150 m , u =-20 cm$
Mirror equation, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$\frac{1}{v}=\frac{1}{f}=\frac{1}{u}a)$
$\frac{1}{v}=\frac{1}{-15}=\frac{1}{-20} \text { Samacheerkalvi.Guide }$
$\frac{1}{v}=\frac{-20-(-15)}{300}=\frac{-5}{300}=\frac{-1}{60}$
$v=-60 cm$The screen is to be placed at distance $60 cm$ to the left of the concave mirror.b) Magnification,
$m =\frac{h^{\prime}}{h}=\frac{-v}{u}$
$m =\frac{h^{\prime}}{h}=\frac{-v}{u}=-3$
As magnification is negative, the image is inverted.

View full question & answer→

Question 45 Marks

A transmission grating has 5000 lines $/ cm$. Calculate the angular separation in second order spectrum of red line $7070 \mathring A$ and blue line $5000 \mathring A$.

Answer

$\text { (i) } \sin \theta= Nm \lambda$
$\sin \theta_R=5 \times 10^5 \times 2 \times 7070 \times 10^{-10}$
$\sin \theta_R=1 / \sqrt{ } 2$
$\theta_R=45^{\circ}$
$\text { (ii) } \sin \theta_B=5 \times 10^5 \times 2 \times 5 \times 10^{-7}$
$=50 \times 10^{-2}=0.5$
$\theta_B=30^{\circ}$
Angular separation $\left(\theta_R-\theta_B\right)=45^{\circ}-30^{\circ}=15^{\circ}$

View full question & answer→

Question 55 Marks

A monochromatic light of wavelength 589 nm is incident on a water surface having refractive index 1.33. Find the velocity, frequency and wave length of light in water, μ (refractive index of water)

Answer

(i)
$
\begin{aligned}
1.33 & =\frac{ V _{\text {air }}}{ V _{\text {medium }}}=\frac{3 \times 10^8}{ V _{ w }} \\
V _{ W } & =\frac{3 \times 10^8}{1.88}=\frac{3}{4} \times 3 \times 10^8 \\
& =2.26 \times 10^8 ms ^{-1}
\end{aligned}
$
(ii)
$
\begin{aligned}
\mu_w & =\frac{\lambda_{\text {air }}}{\lambda_{\text {water }}} \\
\mu_w & =\frac{589 \times 10^{-9}}{1.33} \\
& =442.85 \times 10^{-9} m
\end{aligned}
$
iii)
$
\begin{aligned}
C_w & =V_w / \lambda_w \\
& =\frac{3 \times 10^8}{589 \times 10^{-9}} \\
& 5.093 \times 10^{14} H z
\end{aligned}
$

View full question & answer→

Question 65 Marks

The light of wavelength $590 nm , 596 nm$ are used in turn to study the diffraction taking place a single slit of aperture $2 \times 10^{-1} in$. The distance between the slit and the screen is $1.5 m$. Calculate the separation between the positions of first maximum of the diffraction pattern obtained in the two cases.

Answer

Wave length of first sodium line,$\lambda_1=590 nm$Wavelength of second sodium line,$\lambda_2=596 nm$Aperture or width of single slit, $d =2 \times 10^{-4} m$Distance between the slit and the screen, $D =1.5 m$Separation between the first maximum is given by,$\begin{aligned}x _2- x _1 & =\frac{3}{2} \frac{D \lambda_2}{d}-\frac{3}{2} \frac{D \lambda_1}{d} \\x _2- x _1 & =\frac{3 D}{2 d}\left(\lambda_2-\lambda_1\right) \\x _2- x _1 & =\frac{3 \times 1.5}{2 \times 2 \times 10^{-4}}\left(596 \times 10^{-9}-590 \times 10^{-9}\right) \\& =\frac{3 \times 1.5 \times 6}{4} \times 10^{-5} \\& x _2- x _1=6.75 \times 10^{-5} m\end{aligned}$

View full question & answer→

Question 75 Marks

A parallel beam of monochromatic light is allowed to incident normally on a plane transmission grating having 5000 lines per centimetre. A second order special line is found to be diffracted at an angle $30^{\circ}$ Find the wavelength of the light?

Answer

Data:$N =5000 \text { lines } / cm =5000 \times 10^2 \text { lines } / m$
$m =2 ; \theta=30^{\circ} ; \lambda=?$
$\sin \theta= Nm \lambda$
$\lambda=\frac{\sin \theta}{ Nm }$

View full question & answer→

Question 85 Marks

Young’s double slit experiment two coherent sources of intensity ratio of $64 : 1,$ produce interference fringes. Calculate the ratio of maximum aqd minimum intensities.

Answer

Given data:
$I _1: I _2:: 64: 1 $
$\frac{I_{\max }}{I_{\min }}=? $
$\frac{ I _1}{ I _2}=\frac{ a _1^2}{ a _2^2}=\frac{64}{1}$
$\frac{ a _1}{ a _2}=\frac{8}{1} ; a _1=8 a ^2 $
$\frac{ I _{\max }}{ I _{\min }}=\frac{\left( a + a _2\right)^2}{\left( a - a _2\right)^2}=\frac{\left(8 a _2+ a _2\right)^2}{\left(8 a _2- a _2\right)^2} $
$=\frac{\left(9 a _2\right)^2}{(7 a )^2}=\frac{81}{49} \\{ }_{\max }: I _{\min }:: 81: 49 \text { }$

View full question & answer→

Question 95 Marks

Calculate the Refractive index of the material. Whose polarising angle is $60^{\circ}$.

Answer

$i _{ p }=60^{\circ} $
$\tan i _{ p }= n $
$n =\tan i _{ p }$
$=\tan 60^{\circ} \\n =1.732$

View full question & answer→

Question 105 Marks

A person runs towards a plane mirror at a speed of $1.5 ms ^{-1}$. with what speed does the image approach the person?

Answer

The person moves towards the mirror by $1.5 m$ in one second. Hence the image moves towards the mirror by $1.5 m$ in one second.$\therefore$ The net displacement of the image with respect to the person is $=1.5+1.5=3 m$ per second

View full question & answer→

Question 115 Marks

White light is incident on a small angled prism of angle $5^\circ$ calculate the angular dispersion if the refractive indices of red and violet rays are $1.642$ and $1.656$ respectively.

Answer

Angular dispersion $\delta_v-\delta_r=\left(\mu_v-\mu_r\right) A$
$=(1.656-1.642) 5^{\circ}$
$=0.070^{\circ}$
Angular dispersion $=0.070$

View full question & answer→

Physics Numerical Problems

Test yourself on this topic