CBSE Class 12 Physics Optical Instrument Solved Examples

CBSE Class 12 Physics Optical Instrument Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Optical Instruments – Solved Examples

Example 1. A 35 mm film is to be projected on a 20 m wide screen situated at a distance of 40 m from the film projector. Calculate the distance of the film from the projection lens and the focal length of the projection lens.

Solution. For a projector, the magnification is

\[m=\frac{I}{O}=\frac{v}{u}\]

Taking the magnitudes only,

\[\frac{20\times100\ \text{cm}}{3.5\ \text{cm}}=\frac{40\times100\ \text{cm}}{u}\quad\Rightarrow\quad u=-7\ \text{cm}\]

So the film is 7 cm in front of the projection lens. From the lens formula \(\frac1v-\frac1u=\frac1f\), with \(v=4000\ \text{cm}\) and \(u=-7\ \text{cm}\),

\[\frac{1}{4000}-\frac{1}{-7}=\frac1f\quad\Rightarrow\quad f\approx7\ \text{cm}=70\ \text{mm}\]

because \(\frac1{4000}\ll\frac17\). So the focal length of the projection lens is 70 mm.

Example 2. A compound microscope has an objective of focal length 2 cm and an eyepiece of focal length 5 cm. If an object is placed at a distance of 2.4 cm in front of the field lens, find the magnifying power of the instrument and the length of the tube if (a) the final image is at infinity, (b) the final image is at the least distance of distinct vision (25 cm).

Solution. The object is 2.4 cm in front of the field lens, whose focal length is 2 cm. The field lens will form its image at a distance \(v\) given by

\[\frac1v-\frac1{-2.4}=\frac12\quad\Rightarrow\quad v=12\ \text{cm}\]

so that the magnification of the objective is

\[m_o=\frac vu=\frac{12}{-2.4}=-5\]

(a) Final image at infinity (far point). For the eyepiece, \(\frac1\infty-\frac1{u_e}=\frac15\), so \(u_e=-5\ \text{cm}\) and

\[m_\theta=\frac{D}{f_e}=\frac{25}{5}=5\]\[\text{MP}=m_o\times m_\theta=-5\times5=-25\]

The length of the tube is \(L=v+|u_e|=12+5=17\ \text{cm}\). In this situation the eye is relaxed. For a given microscope the magnifying power is minimum and the length of the tube is maximum.

(b) Final image at the near point (D = 25 cm). For the eyepiece,

\[\frac1{-25}-\frac1{u_e}=\frac15\quad\Rightarrow\quad u_e=-\frac{25}{6}=-4.17\ \text{cm}\]\[m_\theta=1+\frac{D}{f_e}=1+\frac{25}{5}=6\]\[\text{MP}=m_o\times m_\theta=-5\times6=-30\]

The length of the tube is \(L=12+4.17=16.17\ \text{cm}\). In this situation the eye is strained. For a given microscope the magnifying power is maximum and the length of the tube is minimum.

Note: in case (b), \(m_e=\frac{D}{|u_e|}=\frac{25}{25/6}=6=m_\theta\), so \(\text{MP}=m_o\times m_\theta=m_o\times m_e\) is simply the linear magnification.

Example 3. In a compound microscope the objective and the eyepiece have focal lengths of 0.95 cm and 5 cm respectively, and are kept at a distance of 20 cm. The last image is formed at a distance of 25 cm from the eyepiece. Calculate the position of the object and the total magnification.

Solution. The final image is 25 cm from the eyepiece, so for the eyepiece

\[\frac1{-25}-\frac1{u_e}=\frac15\quad\Rightarrow\quad u_e=-\frac{25}{6}\ \text{cm}\]\[m_e=\frac{v_e}{u_e}=\frac{-25}{-25/6}=6\qquad\cdots(1)\]

For the objective, the image distance is

\[v=L-|u_e|=20-\frac{25}{6}=\frac{95}{6}\ \text{cm}\]

If the object is at a distance \(u\) from the objective,

\[\frac{6}{95}-\frac1u=\frac1{0.95}\quad\Rightarrow\quad u=-\frac{95}{94}\ \text{cm}\]

So the object is at a distance \(\frac{95}{94}\) cm in front of the objective. Also

\[m_o=\frac vu=\frac{95/6}{-95/94}=-\frac{94}{6}\qquad\cdots(2)\]

The total magnification is

\[M=m_o\times m_e=-\frac{94}{6}\times6=-94\]

The final image is inverted, virtual, and 94 times the size of the object.

Example 4. A Galilean telescope consists of an objective of focal length 12 cm and an eyepiece of focal length 4 cm. What should be the separation of the two lenses when the virtual image of a distant object is formed at a distance of 24 cm from the eyepiece? What is the magnifying power of the telescope under this condition?

Solution. The object is distant, so \(u=-\infty\) for the objective, and

\[\frac1v-\frac1{-\infty}=\frac1{f_0}\quad\Rightarrow\quad v=f_0=12\ \text{cm}\]

So the objective forms the image \(I_M\) at its focus, 12 cm from \(O\). The eyepiece (focal length \(-4\) cm) forms the final image \(I\) at a distance of 24 cm from it. Therefore

\[\frac1{-24}-\frac1{u_e}=\frac1{-4}\quad\Rightarrow\quad u_e=\frac{24}{5}=4.8\ \text{cm}\]

That is, \(I_M\) is 4.8 cm from the eyepiece \(E\). So the length of the tube is

\[L=OA-EA=12-4.8=7.2\ \text{cm}\]

By definition,

\[\text{MP}=\frac{\theta}{\theta_0}\approx\frac{\tan\theta}{\tan\theta_0}=\frac{AB/EA}{AB/OA}=\frac{OA}{EA}\]\[\text{MP}=\frac{f_0}{u_e}=\frac{12}{4.8}=\frac{10}{4}=2.5\]

So the image is erect and virtual, and it is at a distance of \(24-7.2=16.8\) cm in front of the objective.