Very Short Question Answers

Q1. Define focal length and radius of curvature of a concave lens.

Ans.

Focal length: The distance between the pole and the principal focus of concave lens is called its focal length.

Radius of curvature: The radius of curvature of a concave lens is the radius of the sphere of which the lens is a part.

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Q2. What do you understand by the terms ‘focus’ and‘principal focus’ in the context of lenses?

Ans.

Focus:It is the point at which rays converge or from which they appear to diverge.

Principal focus: It is the point, at which rays parallel to the principal axis converge (in case of a concave lens), or from which they appear to diverge (in case of a concave lens).

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Q3. What is optical density and how it is different from mass density?

Ans.

1. Optical density is a measure of the ability of a substance of transmit the light.
2. Optical density is the ratio of speed of light in the two media.
3. Mass density is the mass per unit volume.

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Q4. What are the laws of reflection through curved mirrors?

Ans.

Law1:The incident, the reflected ray and the normal all lie in the same plane.

Law2:The angle of reflection is always equal to the angle of incidence.

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Q5. Define ‘Power’ of a convex lens. What is its unit.

Ans.

Power of a lens: Reciprocal of focal length of a lens is called power of the lens.

$$ \therefore P=\frac{1}{f}$$

Unit:Diopter (or) m^-1

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Q6. A concave mirror of focal length 10 cm is placed at a distance 35cm from a wall. How far from the wall should an object be placed so that its real image is formed on the wall?

Ans.

Distance  of image from wall $(v)= -35 cm$

focal length $(f)= -10 cm$

$ from \,\, Mirror \,\, formula  \,\,  \frac{1}{u}=\frac{1}{f}-\frac{1}{v} $

$$\therefore  \frac{1}{u}=\frac{1}{-10}-\frac{1}{-35}$$

$$ \frac{1}{u}=\frac{-25}{350} $$

$$ u= -14 cm$$

Distance of Object from Wall $= 35-14=21 cm $

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Q7. A small angled prism of 4^°deviates a ray through 2.48^°Find its refractive index.

Ans.

$A=4^{\circ} \,,\, \: D_{m}=2.48^{\circ}\,\, μ=? $

$ For \, small \, angled \, Prism \,D_{m}=(μ-1)A$

$\therefore \mu=\frac{2.48}{4}+1$

$ \mu= 1.62$

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Q8. What is dispersion? Which color gets relatively more dispersed?

Ans.

Dispersion: The phenomenon of splitting of light into itsconstituentcolours is known as dispersion.

          Violet colour gets more dispersed.

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Q9. The focal length of a concave lens is 30 cm. where an object should placed so that its image is 1/10 of its size?

Ans.

$f=-30\:cm \,,\, \:  Magnification \: m= \frac{1}{10}$

$ For  \, Cancave \, lens \, m=\frac{f}{f+u}$

$\therefore  \frac{1}{10}=\frac{-30}{-30+u}$

$ Object \, distance  = -270\, cm$

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Q10. What is myopia? How can it be corrected?

Ans.

Myopia:The light form a distant object arriving at the eye lens may get converged at a point in front of the retina. This type of defect is called ‘myopia’ or near ‘sightedness’.

It can be corrected by using concave lens.

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Q11.  What is Hypermetropia? How can it be corrected?

Ans.

Hypermetropia: If the eye-lens focusses, the incoming light at a point behind the retina is called

hypermetropia or farsightedness.      

This defect can be corrected by using a convergent lens or convex lens.

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Q12. What focal length should the reading spectacles have for a person for a person for whom the least distance of vision is 50 cm?(The distance of Normal vision is 25 cm.)

Ans.

$\frac{1}{f}=   \frac{1}{v}-\frac{1}{u} $

$\frac{1}{f}=   \frac{1}{-50}-\frac{1}{-25} $

$\therefore \: f= 50\:cm$

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Short Question Answers

Q1. Define critical angle. Explain total internal reflection using a neat diagram.

Ans. Definition: It is an incidence angle  for  which  refraction  angle is equal to  $90^{\circ}$ .

Total internal reflection:  When an incident angle is greater than the critical angle $(i>i_{c})$then the total light is reflected into the same denser medium. This phenomenon is know as Total internal reflection.
1. As the light ray is travelling from denser medium to rarer medium the refracted ray moves away from the normal drawn at the point of incidence
of light ray.
2. As the angle of incidence in denser medium increases the angle of refraction in rarer medium increases. At one particular angle of incidence in denser
medium the refracted light ray just grazes out the interface of two media.For this the angle of incidence in denser medium is called critical angle.
3. When the light ray is travelling from denser to rarer medium and the angle of incidence is greater than critical angle$(i>i_{c})$ then the light ray is reflected back to the same denser medium. This phenomenon is called total internal reflection.

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Q2. Why does the setting sun appear red?

Ans. Due to scattering of light setting sun appear red. 

According to Rayleigh's scattering law the amount of scattering $a\propto\frac{1}{{\lambda}^4}$  (Where $\lambda$-Wave length )

In constitutents colours in the Sun light, Red colour having more wave length than rest of colours.  At sunset or sunrise the sun rays have to pass through a larger distance in the atmosphere.Most of the blue and other short  wavelengths are removed by scattering.The least scattered light i.e., red reaching our eyes. Therefore, the sun looks reddish. 

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Q3. Explain the formation of a rainbow.

Ans. Rainbow:This is a phenomenon due to combined effect of dispersion, refraction and reflection of sunlight by spherical water droplets of rain.

1. Firstly, light refracted into from air to rain droplet.
2.Secondly, refracted ray get reflected in rain droplet internally.
3.Thirdly, after reflection again light refracted into from rain droplet to air , During this light get dispersed.according to order of wavelengths red is deviated less and emerges at angle $42^0$, violet is deviated more and emeges $40^0$.For other colours, angles lie in between these two values.

The secondary rainbow is also formed due to double internal reflection of sunlight in the rain drops.

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Q4. Explain the formation of a mirage

Ans.

1. Mirages are formed due to total internal reflection of light.
2. On hot summer days the air near the ground becomes hotter. Hence it becomes less dense.
3. Light ray from a tall object passes through air with decreasing refractive index towards ground.
4. So, the angle of incidence near the ground exceeds critical angle, as a result total internal reflection takes place.
5. For a distant observer light appears to come from somewhere below ground. So the observer assumes that the light reflected from the ground like by a pool of water.
6. This forms an inverted image of tall object and causes optical illusions to the observers. This phenomenon is known as “mirage”.

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Q5. With a neat labeled diagram explain the formation of image in a simple microscope. 

Ans. Formation of image:The object is adjusted within the principal focus of the convex lens to form the image at the near point. The image is virtual, erect and magnified as shown in figure. The image is on the same side as the object at the least distance of distinct vision.

Equation of linear magnification $$m= (1+\frac{D}{f})$$Working: A simple micro scope consist of a single convex lens fitted in a metal frame.The object is adjusted within the principal focus of the convex lens to form the image at the near point. The image is virtual, erect and magnified as shown in figure. The image is on the same side as the object at the least distance of distinct vision.

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Q6. Define focal length of a concave mirror. Prove that the radius of curvature of a concave mirror is double its focal
length.

Ans. Focal Length: The distance between pole and focus of a concave mirror is focal length of a concave mirror.
R=2f Proof: .Consider a ray parallel to the principal axis striking the mirror at $M$. Then $CM$ will be perpendicular to the mirror at $M$. Let$ θ$ be the angle of incidence, and $MD$ be the perpendicular from $M$ on the principal axis. Then

$\angle MCP=\theta$and$\angle MFP=2\theta$

Now,$\tan\theta=\frac{\mathrm{MD}}{\mathrm{CD}}$ and tan 2$\theta=\frac{\mathrm{MD}}{\mathrm{FD}} \quad--(1)$

For small $\theta$, whtch ts true for paraxtal rays, tan$\theta\approx\theta$,

$\tan2\theta\approx2\theta.$ Therefore,$$\frac{\mathrm{MD}}{\mathrm{FD}}=2\frac{\mathrm{MD}}{\mathrm{CD}}$$$$FD= \frac {CD}2$$Therefore, FD$=f$and CD$=R.$Equatton (9.2) then gives$$f=R/2$$

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Q7. Define Snell’s Law. Using a neat labeled diagram derive an expression for the refractive index of the material of an equilateral prism. 

Ans. Snell’s Law: The ratio of sine of angle of incidence to the sine of angle of refraction is constant.$$\mu=\frac{\sin(i)}{\sin(r)}$$

From quadrilateral AQNR$$\angle A+\angle\mathrm{AQN}+\angle\mathrm{ARN}+\angle{QNR}=360^{\circ}$$$$\angle A+90^{\circ}+ 90^{\circ}+\angle{QNR}=360^{\circ}$$$$\angle A +\angle{QNR}=180^{\circ}$$ From the trlangle QNR,$$r_{_1}+r_{_2}+\angle\mathrm{QNR}=180^{\circ}$$Compartng these two equattons, we get $$r_{_1}+r_{_2}=A\quad-----(1)$$The total deviation $\delta$ is the sum of devlattons at the two faces, $$\delta=(i-r_{_1})+(e-r_{_2})$$$$\therefore \delta=i+e-A\quad-----(2)$$When $\delta = D_{m}$, then $i= e$  and  $r_{l}=r_{2}.$  
from eq(1)and(2)
$2r=$A or $r=\frac{A}{2}$
$D_{\mathrm{m}}=2i-A$,or$i=(A+D_\mathrm{m})/2$
The refractive index of the prism is $$ \mu =\frac{\sin[(A + D_m)/2]}{\sin[A/2]}$$

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