When light travels from an optically denser medium to a rarer medium, why does the critical angle of incidence depend on the color of the light?
Answer
Refractive indices are different for different wavelengths of colour. $\mu=\frac{a+b}{\lambda^2}$, hence the value of critical angle $\sin i_c=\frac{i}{\mu}$ is also different for different colours.
Two thin lenses, whose power is $+5 D$ and $-3 D$. are kept in mutual contact. Find the focal length of the combination.
Answer
According to the question,
$P_1=+5 D$
$P_2=-3 D$
Therefore, $P = P _1+ P _2$
$=+5 D +(-3 D )=+2 D$
Focal length $f=\frac{1}{ P }($ in meter $)$
$=\frac{1}{2}$ meter
$=\frac{1}{2} \times 100$
$=50 \ cm$ Ans.
By looking at the construction of a compound microscope and a telescope, how will you distinguish which type of optical instrument is which?
Answer
The aperture of the objective of a compound microscope is very small and the aperture of the telescope lens is very large. Therefore, the two types of optical instruments can be differentiated by looking at the apertures of the objective.
Any transparent medium surrounded by two surfaces, one of which or both surfaces are spherical is called lens. There are two types of lenses-(1) Convex lens (2) Concave lens. The relationship between the distances of the object, the distance of the image and the focal length of the lens is called lens formula. i.e., $-\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$.
Define the phenomenon of total internal reflection.
Answer
Total Internal Reflection : When the value of the angle of incidence in a denser medium is increased slightly beyond the critical angle, then the entire incident light is reflected. According to the rules, it gets reflected and returns back to the denser medium. This phenomenon is called total internal reflection of light.
(i) Incident ray, refracted ray and the normal separating both the mediums, lie in the same plane. (ii) For any two mediums and for light of a certain colour (wavelength), the result of the sine of the angle of incidence and the sine of the angle of refraction is a constant. If the angle of incidence is $i$ and the angle of refraction is $r$. Then $\sin \frac{\sin i}{\sin r}=$ constant $={ }_1 n_2$.
Write the formula to determine the refractive index of the material of the prism in the case of minimum deviation.
Answer
$n_2=\frac{\sin \left(\frac{ A + D _m}{2}\right)}{\sin \frac{ A }{2}}$ Where ${ }_1 n_2=$ Refractive index of the material of the prism relative to air $A =$ Prism angle $D _m=$ Minimum deviation angle
Magnification (m): The ratio of the size of the image $\left(h^{\prime}\right)$ and the size of the object $(h)$ is called the magnification of the spherical mirror, i.e. $m=\frac{\text { Size of image }\left(h^{\prime}\right)}{\text { Size of object }(h)}$ $\Rightarrow \quad m=\frac{h^{\prime}}{h}=-\frac{v}{u}$
Write the formula of spherical mirror and explain it.
Answer
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$, where $f, v$ and $u$ are the focal length, distance of the image and distance of the object respectively. $f$ is half of the radius of curvature $R$. For a concave mirror, $f$ is negative and for a convex mirror, $f$ is positive.
Write the definition of concave mirror and convex mirror.
Answer
Concave mirror : If the outer part is polished and light is reflected from the inner part, then it is called concave mirror. Convex mirror : If the inner part of a spherical mirror is polished and light is reflected from the outer part, then it is called a convex mirror.
The objective of a telescope is large and the eyepiece is small, whereas the objective of a microscope is small and the eyepiece is large. If a telescope is turned upside down, can it be used like a microscope? Can a microscope be used like a telescope in the same way?
Answer
Not. Because the difference in focal lengths of the lenses of a telescope is much greater than the difference in focal lengths of the lenses of a microscope. In this way, if the telescope is turned upside down and used like a microscope, its magnifying power will be very less. Similarly, if a microscope is turned upside down and used as a telescope, its magnifying power will reduce.
When the last image of the telescope is formed at infinity, then what is the length of its tube? If the image is formed at the minimum distance of clear vision then will the length be less or more than before?
Answer
When the image is formed at infinity then the length of the tube will be $\left(f_o+f_e\right)$. Here $f_o$ is the focal length of the objective lens and $f_e$ is the focal length of the eyepiece is but if the final image is formed at the minimum distance of clear vision then the length of the tube will be ( $f_o+u_e$ ), where $u_e$ is the distance of the image formed by the objective lens from the eyepiece. This length will be less than $\left(f_o+f_e\right)$.
A converging lens and a diverging lens of the same focal length are placed coaxially in contact with each other. Find the focal length and power of the combination.
Answer
If the focal length of the combination is $F,$ then
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f}+\frac{1}{-f}$
$\frac{1}{F}=\frac{1}{f}-\frac{1}{f}=0$
Thus
$F=\frac{1}{0}=\infty$
i.e. $ F =$ Infinite
Therefore, Power $= P =\frac{1}{F}$
$P=\frac{1}{\infty}=0$
Thus, focal length of combined lens is infinite and power of combined lens will be zero.
If we cover the lower half of the reflecting surface of a concave mirror with some opaque (nonreflective) material, then what effect will this have on the image formed by the mirror of an object situated in front of the mirror?
Answer
You can think that half of the image will be visible in the reflection. But assuming that the laws of reflection also apply to the remaining part of the mirror, the mirror will form a complete image of the object. However, since the area of the reflecting surface has reduced, the intensity of the image will reduce, i.e. it will be halved.
How can a bundle of optical fibers be used? Explain.
Answer
Bundles of optical fibers can be used in many ways. Optical fibers are extensively used for transmission and receiving of electrical signals, which are converted into light by suitable transducers. It is clear that optical fibers can also be used for transmitting optical signals. For example, they are used as 'light pipes' for visual observation of internal organs, such as the oesophagus, stomach and intestines.
How will an air bubble behave like a lens in water?
Answer
Both the surfaces of the air bubble are convex. Hence this behaves like a convex lens. But the refractive index of water is greater than the refractive index of air. Therefore, due to the change in the nature of the air bubble present in the water tank, it will act like a concave lens.
The refractive index of the material of a convex lens is 1.5 and the radii of curvature of both its surfaces are equal. Prove that its focal length is equal to the radius of curvature.
Answer
$\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$ $\frac{1}{f}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{-R_2}\right)$ $\Rightarrow \quad \frac{1}{f}=0.5 \times\left(\frac{2}{ R }\right)=\frac{1}{ R }$ or $\quad$$\underline{f=R}$
Can the refractive index of any medium be less than 1 relative to another medium?
Answer
Yes. For example, refractive index of water relative to glass is $ _g n_w=\frac{n_w}{n_g}=\frac{(4 / 3)}{(3 / 2)}=\frac{8}{9} $ which is less than one.
A light ray is refracted from medium $(1)$ to medium $(2).$ Their refractive indices are $n_1$ and $n_2,$ respectively and $n_2 < n_1.$ Write an expression for the critical angle of incidence.
Answer
When angle of incidence $i=$ critical angle $c$ then angle of refraction $r=90^{\circ}$
But according to Snell's law,
$n_1 \sin i=n_2 \sin r$
Therefore,
$n_1 \sin c=n_2 \cdot \sin 90^{\circ}=n_2 \times 1=n_2$
or $ \sin c=\left(\frac{n_2}{n_1}\right) \Rightarrow c=\sin ^{-1}\left(\frac{n_2}{n_1}\right)$
