The relation among u, v and f for a mirror is:
Explanation:
As we know,
$\frac{1}{\text{f}}=\frac{1}{\text{v}}+\frac{1}{\text{u}}$
$\frac{1}{\text{f}}=\frac{\text{u+v}}{\text{uv}}$
$\text{f}=\frac{\text{uv}}{\text{u+v}}$
A point source of light is placed in front of a plane mirror:
Explanation:
A thin lens with focal length f to be used as magnifying glass. Which of the following statements regarding the situation is true?
Explanation:
A converging lens, magnifies the image of an object if it is placed at a distance less than the focal length of the lens.
The phenomena involved in the reflection of radiowaves by ionosphere is similar to:
Solution:
Radio waves are reflected by a layer of atmosphere called the Ionosphere, so they can reach distant parts of the Earth. The reflection of radio waves by ionosphere is due to total internal reflection. It is the same as total internal reflection of light in air during a mirage, i.e., angle of incidence is greater than critical angle.
Important point: The ionized part of the Earth’s atmosphere is known as the ionosphere. Ultraviolet light from the sun collides with atoms in this region knocking electrons loose. The creates ions, or atoms with missing electrons. This is what gives the Ionosphere its name- and it is the free electrons that cause the reflection and absorption of ratio waves.
Which of the following would you prefer to use while reading small letters found in a dictionary?
Explanation:
Convex lens can form an erect, virtual and enlarged image when the object is positioned between its pole and focus.
The instrument that is based on the principle that when an object is placed between first principal focus and the optic centre of convex lens, an upright, virtual and enlarged image on the same side of the object is formed, is:
Explanation:
In astronomical telescope 2 convex lens called eyepiece and objective lens are used and object is placed before eyepiece lens, such that final image inverted, a camera and eye also form inverted image on the screen.
Whereas simple microscope gives an erect, virtual and enlarged image of the object placed between first principal focus and the optic nerve of the convex lens.
In a projector, the image formed is real, inverted magnified on the other side of the lens. This inverted image is again inverted by the film.
One cannot see through fog because .......... :
Explanation:
We cannot see through fog because of scattering.
Atoms and molecules in the air, including anything carried in the air like dust or smoke, will scatter light. Water droplets, as they are present in fog, also scatter light.
The light falling on an object and reflected to a viewer can be scattered to heck and back before it gets to the place where it can be 'seen' by an observer.
So the observer just sees a 'whiteout' instead of being able to make out anything beyond a few meters or so.
The perpendicular distance between the original path of the incident ray and the convergent ray of light coming out of a glass slab is called:
Explanation:
The perpendicular distance between the original path of the incident ray and the emergent ray of light coming out of a glass slab is called lateral displacement. It is proportional to the thickness glass slab.
The phenomenon due to which a ray of light .......... from its path while travelling from one optical medium to another optical medium is called refraction.
Explanation:
Refraction is a phenomenon in which when a ray passes from one medium to another it bends away from its straight-line path due to the difference in optical densities or refractive indices of the two mediums. This bending is known as deviation.
The lateral displacement depends on ______:
Explanation:
Lateral displacement due to a slab is given by $\delta=\text{t}(\mu−1)$; where t is thickness of the medium and μ is refractive index of medium.
prism can produce a minimum deviation $\delta$ in a light beam. If three such prisms are combined, the minimum deviation that can be produced in this beam is:
Explanation:
In combination (refractive angles of prisms reversed with respect to each other), the deviations through two prisms cancel out each other and the net deviation is due to the third prism only.
A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was 4D, the power of a divided lens will be:
Explanation:
Before cut
$\frac{1}{\text{f}}=(\mu-1)\Big(\frac{2}{\text{R}}\Big)=4\text{D}$
After cut
$\frac{1}{\text{f}_1}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_1$
$\&\frac{1}{\text{f}_2}=(\mu-1)\Big(\frac{1}{\text{R}}\Big)=\text{P}_2$
Power of a divided lens will be = P1 + P2
$=(\mu-1)\Big(\frac{2}{\text{R}}\Big)$
$=4\text{D}$
The danger signals installed at the top of tall buildings are red in colour. These can be easily seen from a distance because among all other colours, the red light:
Explanation:
The amount of scattering is inversely proportional to the fourth power of wavelength. And since the wavelength of red has the longest wavelength, the amount of scattering becomes smaller.
Danger signals installed at the top of tall buildings or traffic signals are red so that they can be easily seen from a distance.
The angular dispersion produced by a prism:
Explanation:
If $\mu$ is the average refractive index and A is the angle of prism, then the angular dispersion produced by the prism is given by $\delta=(\mu-1)\text{A}.$
A magnifying glass is used, as the object to be viewed can be brought closer to the eye than the normal near point. This results in;
Solution:
Key concept: A magnifying glass is a single convex lens of lesser focal length.
For magnification when final image is formed at D and $\infty(\text{i.e., m}_\text{D}\text{ and m}_\infty)$.
$\text{m}_\text{D}=\Big(1+\frac{\text{D}}{\text{f}}\Big)_\text{max}\text{ and }\text{m}_\infty=\Big(\frac{\text{D}}{\text{f}}\Big)_\text{min}$
When a magnifying glass is used, the object to be viewed can be brought closer to the eye than the normal near point. This results in a larger angle to be subtended by the object at the eye and hence, viewed in greater detail. Moreover, the formation of a virtual erect and enlarged image takes place.
How can we explain the reddish appearance of sun at sunrise or sunset?
Explanation:
During sunrise or sunset, the light has to pass through greater distance in the atmosphere. The blue light is removed as it gets scattered the most while the red colour is less scattered and reaches the observer. Thus, we find reddish colour of the sun during sunrise or sunset.
An astronomical refractive telescope has an objective of focal length 20m and an eyepiece of focal length 2cm.
Solution:
Key concept:
Used to see heavently bodies.
fobjective > feye lens and dobjective > deye lens.
Intermediate image is real, inverted and small.
Final image is virtual, inverted and small.
Magnification: $\text{m}_\text{D}=-\frac{\text{f}_0}{\text{f}_\text{e}}\Big(1+\frac{\text{f}_\text{e}}{\text{D}}\Big)$ and $\text{m}_\infty=-\frac{\text{f}_0}{\text{f}_\text{e}}$.
Length: $\text{L}_\text{D}=\text{f}_0+\text{u}_\text{e}=\text{f}_0+\frac{\text{f}_\text{e}\text{D}}{\text{f}_\text{e}+\text{D}}$ and $\text{L}_\infty=\text{f}_0+\text{f}_\text{e}$.
The length of the telescope tube is f0 + fe = 20 + (0.02) = 20.02m
Also, $\text{m}=\frac{20}{0.02}=1000$
Also, the image formed is inverted.
When a ray of light passes from a denser to a rarer medium, some part of it gets ....... into the denser medium:
Explanation:
When a ray of light passes from a denser to a rarer medium, some part of it gets refracted into the rarer medium such that it bends away from the normal. Some part of it gets reflected back into the denser medium. The light reflected back into the denser medium is said to be internally reflected.
The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked 1, 2, 3 and 4 (Fig). Which of the four rays correctly shows the direction of reflected ray?
Solution:
The ray PQ of light passes through focus F and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray 2 in the figure.
Important points: We can locate the image of any extended object graphically by drawing any two of the following four special rays:
In image formation from spherical mirrors, only paraxial rays are considered because they:
Explanation:
In Image formation from spherical mirrors, only paraxial rays are considered because they form nearly a point Image of a point source. Angle of Incidence of Paraxial rays is very small.
The critical angle of a material is the angle of incidence for which the angle of refraction is:
Explanation:
critical angle: It is the angle of incidence for which angle of refraction is 90°.
The imaginary line passing through the pole and the center of curvature of the curved mirror is called its _______.
Explanation:
The imaginary line passing through the pole and the center of curvature of the curved mirror is called its principal axis.
The image which can not be taken on the screen is called:
Explanation:
The image which can not be taken on the screen is called a virtual image.
Which of the following will produce greater chromatic aberration?
Explanation:
A lens may be considered as made up of a number of prisms placed one above the other. In a thick lens, the angles of prisms are larger than those in a thin lens. Since, angular dispersion produced by a prism is directly proportional to the angle of prism, a thick lens suffers from greater chromatic aberration.
The optical density of turpentine is higher than that of water while its mass density is lower. Fig. shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Fig, the path shown is correct?
Solution:
Key concept: The Snell's law describes the relation between angle of incidence $\theta_1$ and angle of refraction $\theta_2$:
$\mu\sin\theta_1=\mu_2\sin\theta_2=\text{constant}\ .....(\text{i})$
where $\mu_1$ and $\mu_1$ are refractive indices of the two media.
When a light ray goes from (optically) denser medium to (optically) rarer medium, then it bends away the normal, i.e., $\theta_1<\theta_2$ and vice-versa.
Here, light ray goes from (optically) rarer medium air to optically denser medium turpentine, then it bends towards the normal, i.e., $\theta_1>\theta_2$ whereas when it goes from to optically denser medium turpentine to rarer medium water, then it bends away the normal.
Three transparent media of refractive indices $\mu_1,\mu_2$ and $\mu_3.$ A point object O is placed in the medium $\mu_2.$ If the entire medium on the right of the spherical surface has refractive index $\mu_1,$ the image forms at O'. If this entire medium has refractive index $\mu_3,$ the image forms at O". In the situation shown:
Explanation:
m1, Image is O1
m3, Image is O11
Spherical Surface formula
$\frac{\mu^{11}}{\text{v}}-\frac{\mu_1}{\mu}=\frac{\mu^{11}-\mu^1}{\text{R}}$
If ray goes to m2 to m1 than Image is formed at O1 and if ray goes to m2 to m3 than Image is formed at O11.
A pencil dipped partially in water appears bent because of:
Explanation:
Refraction is a phenomenon in which a light ray incident on a surface separating two transparent media bends at the change of medium.
Snell's law gives the relation between angle of incidence and refraction and the Refractive index of the respective medium.
$\mu_1$sin(i) = $\mu_2$sin(r)
where, i, r are the angles of incidence and refraction respectively.
The image formed by a concave mirror:
Explanation:
| Object | Image |
| O1 | I1 |
| O2 | I2 |
| O3 | I3 |
| O4 | I4 |
Here O4 is virtual oblect & I4 is real image.
When the reflecting or refracting rays do not actually intersect but appear to intersect when produced backwards,
Explanation:
A virtual image is formed when the refracting or reflecting rays do not actually intersect but appear to intersect when they are produced backwards.
In optical instruments, the lenses are used to form images by:
Explanation:
A lens is a piece of glass (plastic) with two refracting surfaces, which are either curved (e.g., a segment of a sphere) or plain.
Lenses are used to form images by refraction in optical instruments (microscopes, telescopes, cameras, etc.)
There are two types of lenses: converging (thickest in the middle) and diverging (thickest at the edges).
Total internal reflection can take place only if:
Explanation:
T.I.R. (Total Internal reflection)
i < C (condition for T.I.R.)
By Snell's Law
$\text{n}_1\sin\text{i}=\text{n}_2\sin90^{\circ}$
$\sin\text{i}=\frac{\text{n}_2}{\text{n}_1}=\sin\text{C}$
$\text{þ} \ \text{C}=\sin^{-1}$
$\because \ -1\leq\sin\text{i}\leq1$
n1 > n2 (so we can conclude that light goes from optically denser medium to rarer medium & incident angle is greater than the critical angle.)
What causes chromatic aberration?
Explanation:
Chromatic aberration occurs when a lens is either unable to bring all wavelengths of color to the same focal plane and/or wavelengths of color are focused at different positions in the focal plane.
The image of an extended object, placed perpendicular to the principal axis of a mirror, will be erect if:
Explanation:
$\text{m}=\frac{-\text{v}}{\mu}=\frac{\text{h}_1}{\text{h}_0}$ (for mirror)
Image will be erect mean height of the object and Image will e lies in same side. It mean if object isreal then Image in virtual. If object is virtula then Image is real.
The angle which subtends the periphery of the spherical mirror at the centre of curvature is called:
Explanation:
distance between extreme points on the periphery of the spherical mirror is called linear aperture and the angle which the periphery of the spherical mirror subtends at the centre of curvature is called angular aperture.
A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black:
Explanation:
By lens formula
$\frac{1}{\text{v}}-\frac{1}{\text{u}}=\frac{1}{\text{f}}$
Due to Black Pointed focal Lenght of the Lens will not change.
So Image will not be shifted due to Black point. But Intensily of Image will decrease.
Which of the following phenomena contributes significantly to the reddish appearance of the sun at sunrise or sunset?
Explanation:
We see red colour of the sun at sunrise or sunset as the sun at horizon and light rays need to travel a greater distance. In the process of scattering, violet, blue and green rays in the original sunlight are removed and the transmitted beam has yellow and red dominant.
The focal length of the objective of a compound microscope if fo and its distance from the eyepiece is L. The object is placed at a distance u from the objective. For proper working of the instrument:
Explanation:
In a compound microscope, the objective lens of a short focal length, fo is used. The focal length of the objective lens is less than the focal length of the eyepiece, fe,. The object is placed at a distance slightly greater than its focal length. The real, inverted image of the object forms somewhere in front of the eyepiece at a distance less than its focal length. This image acts as its object and the final image forms in between length, L of the microscope.
$\therefore$ L > fo + fe > 2fo
Also, L > u
.......... happens when a wave passes from one medium to another at an angle.
Explanation:
Refraction is a phenomenon in which when a ray passes from one medium to another it bends away from its straight-line path due to the difference in optical densities or refractive indices of the two mediums.
The image which can be obtained on a screen is called:
Explanation:
The image which can be obtained on a screen is called a real image.
Which of the following quantities increase when wavelength is increased? Consider only the magnitudes.
Explanation:
The focal length of a lens is inversely proportional to the refractive index of the lens and the refractive index of the lens is inversely proportional to the square of wavelength. Therefore, the focal length is directly dependent on wavelength; it increases when the wavelength is increased.
For reading small letters with a lens:
Explanation:
A concave lens always produces virtual, erect and diminished images and the decrease in the size of the image depends on the position of the object.
Concave lens will shrink the size of the already small letters.
A convex lens produces real and virtual, erect and inverted, diminished, same sized and magnified image of the object, depending upon the position of the object on the principal axis.
When the object is placed between F and 2F of convex lens, an enlarged but inverted image of the object is formed. The magnified image makes it easier to read small letters but the inverted image is undesirable.
When the object is placed at a distance less than the focal length of convex lens, an enlarged and erect image of the object is formed, which makes it easier to read small letters.
An object approaches a convergent lens from the left of the lens with a uniform speed 5m/s and stops at the focus. The image.
Solution:
If an object approaches a convergent lens from the left of the lens with a uniform speed of 5m/s, then the image moves away from lens with a non-uniform acceleration.
Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because:
Solution:
Key concept: The light from the pencil is refracted when it passes from the water into air, bending away from the normal as it moves from high to low refractive index.
When light from the submerged object before reaching to the observer gets, refracted from water surface, the rays bend away from normal and the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air. Also the apparent depth of the .points close to the edge are nearer the surface of the water compared to the points away from the edge.
As we move towards right, the angle of incident increases and becomes equal to critical angle. Hence some of the points of the object far away from the edge may not be visible because of total internal reflection.
If a ray of light goes from a rarer medium to a denser medium, will it bend towards the normal or away from it?
Explanation:
Refraction is the bending of light rays after entering a medium where its speed is different. Due to refraction of light, when a ray of light passes from a rarer medium to a denser medium, bends towards the normal to the boundary between the two media. The amount of bending depends on the indices of refraction of the two media. Hence, when a ray of light from air enters a denser medium, it bends towards the normal.
You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is 90º. Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
Solution:
Key Concept: According to Couchy relationship,
$\lambda\propto\frac{1}{\mu}$
Smaller the wavelengh higher the refractive index and consequently smaller the critical angle.
We know $\text{v}=\text{f}\lambda$ the frequency of wave remains unchanged with medium hence $\text{v}\propto\lambda$.
The critical angle, sin $\text{C}=\frac{1}{\mu}$
Also, velocity of light, $\text{v}\propto\frac{1}{\mu}$
According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength. As $\lambda_{\text{blue}}<\lambda_\text{yellow}$ hence $\text{v}_{\text{blue}}<\text{v}_\text{yellow}$, it means $\mu_{\text{blue}}<\mu_\text{yellow}$.
It means critical angle for blue is less than yellow colour, the critical angle is least which facilitates total internal reflection for the beam of blue light.
The apparent vertical shift of the image of a coin placed at the bottom of a water tank having constant depth of water is proportional to (given refractive index of water = $\mu$).
Explanation:
The apparent vertical shift is given by the equation $\text{AD}=\frac{\text{RD}}{\mu}$, that is, $\mu=\frac{\text{RD}}{\text{AD}}$, where AD is the apparent vertical shift, RD is the real depth of the tank and n is the refractive index of the denser medium, that is, water.
As real depth and refractive index is going to be constant always, the apparent shift or depth will be independent of viewing angle.
From the above equation, we can see that the apparent vertical shift is inversely proportional to the refractive index. That is, $\frac{1}{\mu}$.
A convex lens is made of a material having refractive index 1.2. Both the surfaces of the lens are convex. If it is dipped into water $(\mu=1.33),$ it will behave like:
Explanation:
Here P, P1 & P2 are the Power of Lenses.
P = P1 + P2
$\frac{1}{\text{f}}=\frac{1}{\text{f}_1}=\frac{1}{\text{f}_2}$
$(\mu-1)\Big(\frac{2}{\text{R}}\Big)+(\mu'-1)\Big(\frac{-1}{\text{R}}\Big)$
$(1.2-1)\Big(\frac{2}{\text{R}}\Big)-\Big(\frac{4}{3}-1\Big)\Big(\frac{1}{\text{R}}\Big)$
$\frac{1}{\text{f}}=\frac{2}{5\text{R}}-\frac{1}{3\text{R}}$
$\frac{1}{\text{f}}=\frac{6-5}{15\text{R}}$
$\text{f}=15\text{R}$
Focal lenght of combined is positive, but it's magnitude in capair to f1 & f2 is High. So it will be hare like a divergent lens.
In case of a real and inverted image, the magnification created by the mirror is:
Explanation:
Magnification of image created by mirror is defined as
$\text{m}=\frac{\text{size of object}}{\text{size of image}}$
and in case of inverted image. Size of image is negative whereas size of object is positive. Hence , magnification produced is negative and it can be unity when object is placed at center of curvature and infinity when object is at focus.
If the light moving in a straight line bends by a small but fixed angle, it may be a case of:
Explanation:
When the light strikes on a surface nearly parallel to it, it then bends by a small and fixed angle after reflection. Also, when the light travels from one medium to another with slight differences in their refractive indices, it bends by a small angle. Thus, the bending of light by a small but fixed angle can be the case of either reflection or refraction.
Shift by which the object appears to be raised depends on:
Explanation:
Shift by which the object appears to be raised depends on these 4 factors. The shift increases with increase in refraction index of medium and also with the thickness of denser medium.