List four characteristics of the images formed by plane mirrors.


Characteristics of the images formed by plane mirrors are:

i) The image formed is of the same size as that of the object.

ii) Images are formed behind the mirrors and are at the same distance from the mirror as that of the object.

iii) The virtual and erect image is formed.

iv) The images are laterally inverted. 

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(a) If the image formed by a lens is diminished in size and erect, for all positions of the object, what type of lens is it?

(b) Name the point on the lens through which a ray of light passes undeviated.

(c) An object is placed perpendicular to the principal axis of a convex lens of focal length 20 cm. The distance of the object from the lens is 30 cm. Find (i) the position (ii) the magnification and (iii) the nature of the image formed.


(a) If the image formed by a lens is diminished in size and erect, for all positions of the object, then the lens is a concave lens.

(b) The point on the lens through which a ray of light passes undeviating is known as Pole.

(c) Given,

Object distance, u = -30 cm

Focal length, f =20 cm

i) Now, using the len’s formula,

 1 over straight f space equals 1 over straight v minus 1 over straight u

1 over straight v equals 1 over straight f plus 1 over straight u

straight v equals fraction numerator straight u space xf over denominator straight u plus straight f end fraction

straight v equals fraction numerator left parenthesis negative 30 right parenthesis straight x 20 over denominator 20 minus 30 end fraction
straight v equals space 60 space cm

The space image space is space formed space at space straight a space distance space of space 60 cm space of space the space other space side space of space the space optical space centre.
ii right parenthesis space Magnification comma space straight m equals straight m equals space minus straight v over straight u space equals negative fraction numerator 60 over denominator negative 30 end fraction space equals 2

 (iii) Image formed is inverted.

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One-half of a convex lens is covered with a black paper. Will such a lens produce an image of the complete object? Support your answer with a ray diagram.

(b) An object 5 cm high is held 25 cm away from a converging lens of focal length 10 cm.

(i) Draw the ray diagram and

(ii) Calculate the position and size of the image formed.

(iii) What is the nature of the image?


(a) Image is formed by a large number of rays from the object. If one part of the lens is blackened, image will be formed. But, intensity of the image will be reduced.

 

(b) i)  

 

 

Object distance, u = -25 cm

Focal length, f = 10 cm

Height of the image, h = 5 cm

ii) Now, using the lens formula,
1 over straight f equals 1 over straight v minus 1 over straight u
we space have comma
1 over straight v space equals 1 over straight f plus 1 over straight u

straight v equals space fraction numerator straight u space straight x space straight f over denominator straight u plus straight f end fraction

straight v equals fraction numerator negative 25 space straight x space 10 over denominator 10 minus 25 end fraction space equals 16.67 space cm

Now comma space magnification
straight m equals negative straight v over straight u space equals negative straight h subscript straight i over straight h subscript 0
equals space straight h subscript straight i space equals fraction numerator straight v space straight x space straight h subscript 0 over denominator straight u end fraction space equals fraction numerator 6.67 space straight x space 5 over denominator left parenthesis negative 25 right parenthesis end fraction equals negative 3.34 space cm

(iii) Negative sign indicates that the image is real & inverted. 

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A convex lens can form a magnified erect as well as a magnified inverted image of an object placed in from of it. Draw ray diagram to justify this statement stating the position of the object with respect to the lens in each case.

An object of height 4 cm is placed at a distance of 20 cm from a concave lens of focal length 10 cm. Use lens formula to determine the position of the image formed.


a) When the object is placed between O and F1, magnified erect image is formed.

                               

b) When the object is placed between F1 and 2F1 , magnified inverted image is formed.

                    

Object distance, u = 20 cm

Image distance, v=?

Focal length, f = 10 cm

According to the sign convention, f = -10 cm and u=-20 cm

Now, using the lens formula,

 1 over straight f space equals 1 over v minus 1 over u

1 over v space equals 1 over f plus 1 over u

1 over v space equals fraction numerator 1 over denominator negative 10 end fraction plus fraction numerator 1 over denominator negative 20 end fraction

equals fraction numerator negative 3 over denominator 20 end fraction
equals space v equals fraction numerator negative 20 over denominator 3 end fraction space equals negative 6.6 space c m

The image is formed at a distance of 6.6 cm from the lens at the same side where the object is placed.

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(a) State the laws of refraction of light. Explain the term absolute refractive index of a medium and write an expression to relate it with the speed of light in vacuum.

(b) The absolute refractive indices of two media 'A' and 'B' are 2.0 and 1.5 respectively. If the speed of light in medium 'B' is , calculate the speed of light in:

(i) vacuum,

(ii) medium 'A'.


Laws of refraction states that:

(1) The incident ray, the refracted ray and the normal to the interface of two media at the point of incidence all lie in the same plane.
(2)  For the light of a given color and for given pair of media, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant.

This is also known as Snell's Law.

Mathematically it can be written as:

fraction numerator sin space straight i over denominator sin space straight r end fraction space equals space c o n s tan t space equals space mu presuperscript A subscript B

Here, straight mu is the refractive index of medium B with respect to medium A.

Refractive index of a medium with respect to the vacuum is known as the absolute refractive index.

 straight mu subscript straight B space equals straight c over straight v semicolonc is the speed of light in vacuum and c is the speed of light in medium B.

b) Absolute refractive of medium A, straight mu subscript straight A =2   

Absolute refractive index of medium B,  straight mu subscript straight B =1.5

 

straight i right parenthesis space For space medium space straight B comma
1.5 space equals space fraction numerator straight c over denominator 2 space straight x space 10 to the power of 8 end fraction

rightwards double arrow space straight c space equals space 2 space straight x 1.5 space straight x space 10 to the power of 8

rightwards double arrow space straight c equals space 3 space straight x space 10 to the power of 8

Speed space of space light space in space vaccum space is space 3 space straight x space 10 to the power of 8

ii right parenthesis space For space medium space straight A comma
space straight mu subscript straight A space equals straight c over straight v
rightwards double arrow space straight v equals straight c over μA

straight v equals fraction numerator space 3 space straight x space 10 to the power of 8 over denominator 2 end fraction
space equals 1.5 space straight x space 10 to the power of 8 space is space the space required space speed space of space light space in space medium space straight A comma.

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(a) State the laws of refraction of light. Give an expression to relate the absolute refractive index of a medium with speed of light in vacuum.                                 

(b) The refractive indices of water and glass with respect to air are 4/3 and 3/2 respectively. If the speed of light in glass is 2 × 108 ms−1, find the speed of light in (i) air, (ii) water. 

(a) Laws of refraction states that:

First law of refraction: The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This is known as Snell’s law.

Mathematically, we have  

sini over sinr space equals space n subscript a b end subscript


Here, nab is the relative refractive index of medium a with respect to medium b.

Second law of refraction: The incident ray, the refracted ray, and the normal to the interface of two media at the point of incidence; all lie in the same plane.

If the light ray goes from medium 1 to 2 then the refractive index of medium 1 with respect to medium 2 is,

straight n subscript 12 space equals space fraction numerator Speed straight space of straight space light straight space in straight space medium space 2 over denominator speed straight space of straight space light straight space in straight space medium space 1 end fraction equals straight v subscript 2 over straight v subscript 1
where v1 and v2 are the speeds of light in medium 1 and 2 respectively.

Given,

Refractive index of water, nw = 4/3, and

Refractive index of glass, ng= 3/2

Speed of light in glass, vg = 2×108 m/s 
i.  straight n subscript straight g over straight n subscript straight a equals v subscript a over v subscript g space  where, na is the refractive index of light in air and va is the speed of light in air. 

space space space v subscript a space equals straight n subscript straight g over straight n subscript straight a cross times v subscript g
equals fraction numerator bevelled italic 3 over italic 2 over denominator italic 1 end fraction cross times left parenthesis 2 cross times 10 to the power of italic 8 right parenthesis
equals space 3 cross times 10 to the power of 8 space m divided by s 
is the required speed of light in air.

ii. Speed of light in water is given by,

straight n subscript straight g over straight n subscript straight w space equals space v subscript w over v subscript g space comma space W h e r e space v subscript w space equals space S p e e d space o f space l i g h t space i n space w a t e r

T h e r e f o r e comma space
v subscript w space equals space n subscript g over n subscript w space cross times space v subscript g space equals space end subscript space fraction numerator 3 divided by 2 over denominator 4 divided by 3 end fraction space cross times space left parenthesis 2 space cross times space 10 to the power of 8 space end exponent right parenthesis space equals space 2.25 space cross times space 10 to the power of 8



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