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light - reflection and refraction

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CBSE Gujarat Board Haryana Board

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Class 10 Class 12
What is light? Mention its two important uses.

Light: Light is a form of electromagnetic radiation (radiant energy) which makes things visible. When the light rays falls on objects, it is reflected back and enters our eyes. This produces the sensation of vision and hence, we are able to see the objects around us. 

Important uses of light

1. Light enables us to see even through a transparent medium because light is transmitted through it. 

2. Light makes things around us visible. 

2. Light is highly useful in modern communication. We can transmit thousands of telephonic conversations simultaneously via optical fibres over long distances.

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A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.


Focal length, f = - 15 cm    [f is - ve for a concave lens]
Image distance, v = - 10 cm [Concave lens forms virtual image on same side as the object, so v is - ve]
As,                            1 over straight f space equals space 1 over straight v minus 1 over straight u
therefore               1 over straight u space equals space 1 over straight v minus 1 over straight f space equals space fraction numerator 1 over denominator negative 10 end fraction minus fraction numerator 1 over denominator negative 15 end fraction space equals space fraction numerator negative 3 plus 2 over denominator 30 end fraction space equals negative 1 over 30
               Object distance, u = -30 cm. Ans. 

Drawing the ray diagram: Using a scale of 1: 5, we get v = - 2 cm, f = - 3 cm. We draw the ray diagram as follows:
(i) Draw the principal axis (a horizontal line).
(ii) Draw a convex lens, keeping principal centre (C) on the principal axis.
(iii) Mark points F and B on the left side of lens at a distance of 3 cm and 2 cm respectively.
(iv) Join any point D (nearly at the top of lens) and F by a dotted line.
(v) Draw a line AD, parallel to principal axis.

Focal length, f = - 15 cm    [f is - ve for a concave lens]Image d

(vi) Draw a line A'B', perpendicular to principal axis from B'.
(vii) Draw a line CA', backwards, so that it meets the line from D parallel to principal axis at A.
(viii) Draw a line AB, perpendicular (downwards) from A to meet the principal axis at B.
(ix) The AB is position of object. Measure distance BC. It will be found to be equal to 6 cm.
Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens.



445 Views

An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.

Converging lens means a convex lens. As the distances given in the question are large, so we choose a scale of 1: 5, i.e., 1 cm represents 5 cm. Therefore, on this scale 5 cm high object, object distance of 25 cm and focal length of 10 cm can be represented by 1 cm high, 5 cm and 2 cm lines respectively. Now, we draw the ray diagram as follows:
(i) Draw a horizontal line to represent the principal axis of the convex lens.
(ii) Centre line is shown by DE.
(iii) Mark two foci F and F' on two sides of the lens, each at a distance of 2 cm from the lens.
(iv) Draw an arrow AB of height 1 cm on the left side of lens at a distance of 5 cm from the lens.
(v) Draw a line AD parallel to principal axis and then, allow it to pass straight through the focus (F') on the right side of the lens.
(vi) Draw a line from A to C (centre of the lens), which goes straight without deviation.
(vii) Let the two lines starting from A meet at A'.
(viii) Draw AB', perpendicular to the principal axis from A'.
(ix) Now AB', represents the real, but inverted image of the object AB.
(x) Then, measure CB' and A'B'. It is found that CB' = 3.3 cm and A'B' = 0.7 cm.

Converging lens means a convex lens. As the distances given in the qu

(xi) Thus the final position, nature and size of the image A'B' are:
         (a) Position of image A'B' = 3.3 cm × 5 = 16.5 cm from the lens on opposite side.
         (b) Nature of image A’B’: Real and inverted. 
         (c) Height of image A'B': 0.7 × 5 = 3.5 cm, i.e., image is smaller than the object.



 
586 Views

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

We are given a convex mirror.

Here, we have 

Object size, h = + 5 cm 
Object distance, u = -20 cm
Radius of curvature, R = + 3.0 cm [R is +ve for a convex mirror]
 Focal length ,  f = R2 = +15 cm 

From mirror formula,

                      1v = 1f-1u 

we have, 

                      1v= 1+15-1-20      = 4+360     = 760 

Image distance, v = 607 8.6 cm. 

Magnification, m = -vu= h'h 
Therefore, 

 Image size, h' = -vhu                          = -8.6 × 5-20                         = 2.15  2.2 cm. 

A virtual and erect image of height 2.2 cm is formed behind the mirror (because v is positive) at a distance of 8.6 cm from the mirror. 

1124 Views

A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.


Focal length, f = - 15 cm    [f is - ve for a concave lens]
Image distance, v = - 10 cm [Concave lens forms virtual image on same side as the object, so v is - ve]
As,                                    1 over straight f space equals 1 over straight v minus 1 over straight u

therefore         1 over straight u space equals 1 over straight v minus 1 over straight f space equals space fraction numerator 1 over denominator negative 10 end fraction minus fraction numerator 1 over denominator negative 15 end fraction space equals space fraction numerator negative 3 plus 2 over denominator 30 end fraction space equals space minus 1 over 30
              Object distance, straight u space equals space minus 30 space cm. space space space Ans. space

Drawing the ray diagram: Using a scale of 1: 5, we get v = - 2 cm, f = - 3 cm. We draw the ray diagram as follows:
(i) Draw the principal axis (a horizontal line).
(ii) Draw a convex lens, keeping principal centre (C) on the principal axis.
(iii) Mark points F and B on the left side of lens at a distance of 3 cm and 2 cm respectively.
(iv) Join any point D (nearly at the top of lens) and F by a dotted line.
(v) Draw a line AD, parallel to principal axis.


Focal length, f = - 15 cm    [f is - ve for a concave lens]Image d

(vi) Draw a line A'B', perpendicular to principal axis from B'.
(vii) Draw a line CA', backwards, so that it meets the line from D parallel to principal axis at A.
(viii) Draw a line AB, perpendicular (downwards) from A to meet the principal axis at B.
(ix) The AB is position of object. Measure distance BC. It will be found to be equal to 6 cm.
Thus, object is placed at a distance of 6 cm × 5 = 30 cm from the lens.





1074 Views

An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed so that a sharp focussed image can be obtained? Find the size and the nature of the image.

We are given a concave mirror. 

Here, 
Object size, h = + 7.0 cm
Object distance, u = - 27 cm
Focal length, f = - 18 cm 
Image distance, v = ?

Image size, h' = ? 

Now, using the mirror formula, 

                     1u+1v = 1f

                  1v = 1f-1u 

                      = 1-18-1-27 = -3+254 = -154

i.e.,                  v = -54 cm 

The screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image. 

Magnification, m = h'h = -vu 

 Image size, 
                        h' = -vhu   

                           =-(-54)×(+7)(-27)= -14 cm 

The image is real, inverted and enlarged in size.

1241 Views