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Ray Optics and Optical Instruments — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsRay Optics and Optical Instruments5 Marks
Question
$(a)$ What is meant by compound microscope? Draw a ray diagram of image formation by a compound microscope. Briefly describe its working and derive the formula for its total magnification.
$(b)$ How does the resolving power of a microscope change when :
$(i)$ The diameter of the objective lens decreases,
$(ii)$ The wavelength of incident light increases? Justify your answer in each case.
✓

Answer

$(a)$ Compound microscope : It is an optical instrument made of two lenses which forms a magnified image of a nearby microscopic object. The angle of view formed by this image on the observer's eye is very large, as a result the eye can see the big object.
Ray diagram of image formation by compound microscope :
Image
Image
Working method:
As shown in the figure, the object $AB$ is placed in front of the objective lens at a distance beyond the focal length of the lens.
The rays emanating from the object get refracted through this lens and form $A _1 B_1$ real, inverted and magnified images of the object. This image acts as an object for the eye lens.
Now we adjust the eye lens in such a way that images $A _1 B_1$ are formed between the focus of the eye lens and the center of light.
The eye lens actually functions as a simple microscope and now its magnified virtual image $A _2 B_2$ is formed.
The position of the eye lens is adjusted such that the final image $A _2 B_2$ formed at a distance $D$ from the eye $A _2 B_2$ is erect with respect to the images $A _1 B_1$ but inverted with respect to the object $AB$ .
Magnifying power :
​​​​​​​The magnifying power of a compound microscope is the ratio of the angle of view formed on the eye by the final image formed by it and the angle of view formed on the eye by the object $($when the object is at the minimum is equal to distance of clear vision and is being seen directly without the instrument$)$.
Hence magnifying power i.e. angular magnification
$m=\frac{\text { The angle made by the final image at the eye }}{\text { The angle made by the object at the eye }}$
$($when the object is at distance $D)$
Suppose the last image $A _2 B_2$ make an angle $B$ at eyepiece $E$. Since the eye is close to the eyepiece, hence $\beta$ can be considered as the angle formed by $A _2 B_2$ on the eye $($Fig. $I)$. If the object $AB$ is placed at the minimum distance $D$ of clear vision from the eye and makes an angle $\alpha$ at the eye when held straight by the eye $($Fig. $II),$ then the magnifying power of the microscope
$m=\frac{\beta}{\alpha} \ldots(1)$
Since the object is small, the angles $\alpha, \beta$ will also be very small.
Therefore, in their place we can take their tangent $(\tan)$
Therefore $\alpha=\tan \alpha$ and $\beta=\tan \beta$
$\therefore m=\frac{\tan \beta}{\tan \alpha} \ldots(2)$
But $($Fig. $II)$ from right angled triangle ABE,
$\tan \alpha=\frac{A B}{D}\ldots(3)$
and $($Fig. $I)$ from right angled triangle $B _1 A_1 E$,
$\tan \beta=\frac{A_1 B_1}{EA_1}\ldots(4)$
After putting the values of equation $(3)$ and $(4)$ in equation $(2),$
$ m=\frac{A_1 B_1 / EA_1}{AB / D}=\frac{A_1 B_1}{AB}\left(\frac{D}{EA_1}\right) \ldots(5)$
If the distances of the object AB and the images $A _1 B_1$ from the objective $O$ are $u_0$ and $v_0$ respectively, then taking their proper signs,
$\frac{ A _1 B_1}{ AB }=\frac{+v_0}{-u_0}$
$\left(\because \frac{\text { Size of the image }}{\text { Size of the object }}=\frac{\text { Distance of the lens from } \\ \text { image }}{\text { Distance of the object from } \\ \text { the lens }}\right)$
If the distance of $A _1 B_1$ from the eyepiece $E\ ($for this, the virtual object$)$ is $u_{ e }$ then with appropriate sign
$EA _1=-u_e$
$\therefore$ Putting these values in equation $(5)$ and taking $D$ as negative,
$m=\frac{+v_0}{-u_0}\left(\frac{- D }{-u_e}\right)$
or $ m=\left[\frac{v_0}{u_0}\left(\frac{D}{u_e}\right)\right] \ldots(6)$
$(b)$ The resolving power of the microscope is $\propto \frac{n \sin \theta}{\lambda}$ where $\theta$ a half angle of the cone.
$(i)$ If the diameter of the objective lens decreases then the resolving power of the microscope also decreases, since the resolving power is directly proportional to the diameter.
$(ii)$ Resolving power $\propto \frac{1}{\lambda}$.
This means that as the value of wavelength $(\lambda)$ increases, the resolution power of that decreases.

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