An object 2 cm high is placed at a distance of 64 cm from the screen. On placing a convex lens 32 cm from the object it is found that a distinct image of the object is formed on the screen. What is the focal length of the convex lens and size of the image formed on the screen? Draw a ray diagram to show the formation of the image in this position of the object with respect to the lens.

Here,  h = + 2 cm,   u = -32 cm
          v = 64 - 32 = + 32 cm                 [v is +ve for a real image]
Now,  1 over straight f space equals space 1 over straight v minus 1 over straight u space equals space 1 over 32 minus fraction numerator 1 over denominator negative 32 end fraction space equals space 1 over 32 plus 1 over 32 space equals 2 over 32 space equals 1 over 16
therefore         f = + 16 cm Ans. 
Magnification,   m = fraction numerator straight h apostrophe over denominator straight h end fraction space equals straight v over straight u
therefore  Image size,  straight h apostrophe space equals space vh over straight u space equals space fraction numerator left parenthesis plus 32 right parenthesis space left parenthesis plus 2 right parenthesis over denominator left parenthesis negative 32 right parenthesis end fraction space equals space minus 2 space cm. space Ans. space
The negative sign shows that the image is real and inverted.
The ray diagram showing the formation of image for this case is shown below:

Here,  h = + 2 cm,   u = -32 cm          v = 64 - 32 = + 32 cm


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The ratio of the focal length of spherical mirror to its radius of curvature is
  • 0.5

  • 1

  • 2

  • 3


A.

0.5

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By stating the sign-convention and assumptions used, derive the relation between object distance u, image distance v and focal length f for a thin convex lens, when it forms real image of an object of finite size.

New cartesian sign convention for refraction of light through spherical lenses. According to this sign convention:
(i) All distances are measured from the optical centre of the lens.
(ii) The distances measured in the same direction as the incident light are taken positive.
(iii) The distances measured in the direction opposite to the direction of incident light are taken negative.
(iv) Heights measured upwards and perpendicular to the principal axis are taken positive.
(v) Heights measured downwards and perpendicular to the principal axis are taken negative.

New cartesian sign convention for refraction of light through spheric
Assumptions used in the derivation of lens formula:
(i) The lens used is thin.
(ii) The aperture of the lens is small.
(iii) The incident and refracted rays make small angles with the principal axis.
(iv) The object is a small object placed on the principal axis.
Derivation of lens formula for a convex lens. As shown in Fig, consider an object AB placed perpendicular to the principal axis of a thin convex lens between its F1 and 2F1. A real, inverted and magnified image AB' is formed beyond 2F2 on the other side of the lens.

New cartesian sign convention for refraction of light through spheric
Fig. Real image formed by a convex lens.
increment straight A apostrophe straight B apostrophe straight O space space and space increment space ABO space are space similar comma
therefore                                      fraction numerator straight A apostrophe straight B apostrophe over denominator AB end fraction space equals space fraction numerator OB apostrophe over denominator BO end fraction                                 ...(1)
Also,  increment space straight A apostrophe straight B apostrophe straight F space space space and space increment space MOF  are similar,
therefore                                  fraction numerator straight A apostrophe straight B apostrophe over denominator MO end fraction space equals space fraction numerator FB apostrophe over denominator OF end fraction
But MO = AB,
therefore                                    fraction numerator straight A apostrophe straight B apostrophe over denominator AB end fraction space equals space fraction numerator FB apostrophe over denominator OF end fraction                                      ...(2)

From (1) and (2), we get   fraction numerator OB apostrophe over denominator BO end fraction space equals fraction numerator FB apostrophe over denominator OF end fraction space equals space fraction numerator OB apostrophe space minus space OF over denominator OF end fraction
Using new Cartesian sign convention, we get
                    Object distance,    BO = -u
                   Image distance, OB' = +v
                   Focal length, OF= + f
therefore                             fraction numerator straight v over denominator negative straight u end fraction space equals space fraction numerator straight v minus straight f over denominator straight f end fraction
Derivation of lens formula is not included in CBSE syllabus.
or                 vf space equals space minus uv space plus space uf
or                  uv space equals space uf space minus space vf
Dividing both sides by uvf, we get
                           1 over straight f space equals space 1 over straight v minus 1 over straight u
This proves the lens formula for a convex lens.

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Draw a ray diagram to show the formation of the image of an object placed between f and 2f of a thin concave lens. Deduce the relation between the object distance, the image distance and the focal length of the lens under this condition.

Derivation of lens formula for a concave lens: The ray diagram for the formation of the image of an object placed between F1 and 2F1 of a thin concave lens is shown in Fig. The image A'B' is virtual and erect.


Derivation of lens formula for a concave lens: The ray diagram for th
increment straight A apostrophe straight B apostrophe straight O space and space increment ABO are similar,
 therefore                                          fraction numerator straight A apostrophe straight B apostrophe over denominator AB end fraction equals space fraction numerator OB apostrophe over denominator OB end fraction                             ...(1)
 increment straight A apostrophe straight B apostrophe straight F space space and space increment space OMF  are similar,
 therefore                                      fraction numerator straight A apostrophe straight B apostrophe over denominator MO end fraction space equals space fraction numerator straight B apostrophe straight F over denominator OF end fraction
But MO = AB,
therefore                                       fraction numerator straight A apostrophe straight B apostrophe over denominator AB end fraction space equals space fraction numerator straight B apostrophe straight F over denominator OF end fraction                                  ...(2)
From (1) and (2), we get          fraction numerator OB apostrophe over denominator OB end fraction space equals fraction numerator straight B apostrophe straight F over denominator OF end fraction space equals fraction numerator OF minus OB apostrophe over denominator OF end fraction
According to new cartesian sign convention,
                        Object distance,  OB = -u
                         Image distance,  OB' = -v
                           Focal length, OF = -f
Derivation is not included in CBSE syllabus.
therefore                                     fraction numerator negative straight v over denominator negative straight u end fraction space equals space fraction numerator negative straight f plus straight v over denominator negative straight f end fraction
or                                  vf space equals space uf space minus space uv
or                                   uv space equals space uf space minus space vf
Dividing both sides by uvf, we get
                                    1 over straight f space equals space 1 over straight v minus 1 over straight u
This is the required relation between u, v and f for a concave lens.





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Draw ray diagrams to show the formation of a three times magnified
(i) real image
(ii) virtual image of an object kept in front of a converging lens. Mark the positions of object, F, 2F, O and position of image clearly in the diagram.
An object of size 5 cm is kept at a distance of 25 cm from the optical centre of a converging lens of focal length 10 cm. Calculate the distance of the image from the lens and size of the image.


(i) The ray diagram for three times magnified real image is shown below:

(i) The ray diagram for three times magnified real image is shown bel

(ii) The ray diagram for three times magnifield virtual image is shown below.

(i) The ray diagram for three times magnified real image is shown bel
Here, h = + 5 cm,   u = - 25 cm,   f = + 10 cm,  v = ?,  h = ?
From lens formula,
               1 over straight v space equals space 1 over straight f plus 1 over straight u space equals space 1 over 10 plus fraction numerator 1 over denominator negative 25 end fraction space equals space fraction numerator 5 minus 2 over denominator 50 end fraction space equals space 3 over 50

or                        straight v space equals space 50 over 3 space cm space equals space 16.67 space cm. space space Ans. space

Also,                 straight m space equals space fraction numerator straight h apostrophe over denominator straight h end fraction space equals space straight v over straight u

therefore           fraction numerator straight h apostrophe over denominator plus 5 space cm end fraction space equals space fraction numerator 50 divided by 3 over denominator negative 25 end fraction space equals space minus 2 over 3

or                      straight h apostrophe space equals space minus fraction numerator 2 cross times 5 over denominator 3 end fraction cm space equals space minus 3.33 space cm. space Ans. space


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