Question

Two lenses of power \[ + 10D\] and \[ - 5D\] are placed in contact. Where an object should be held from the lens, so as to obtain a virtual image of magnification \[2\]?
\[
  A. 5cm \\
  B. - 5cm \\
  C. 10cm \\
  D. - 10cm \\
\]

Answer 1

Hint: Power of a lens is reciprocal of its focal length. For a combination of lenses, the total power is just the addition of the individual powers of constituent lenses. Magnification is the ratio of image height to the object height and it is equal to the ratio of image distance to the object distance. If the produced image is virtual then the magnification must have a positive value.

Formula used:
Object distance for lenses is always negative.
The lens formula connecting the object distance u, image distance v and focal length f is given by:
\[\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}\]
Magnification produced by the lens is given by:
\[m = \dfrac{v}{u} = \dfrac{{{h_2}}}{{{h_1}}}\]
Where $h_1$ is the object height and $h_2$ is the image height.
Image will be virtual and erect if the magnification has a positive value.
Image will be real and inverted if the magnification has a negative value.
The relation between focal length f of a lens and power P is given by
\[P = \dfrac{1}{f}\]
The power of two combination of lenses having individual powers $P_1$ and $P_2$ is:
\[P = {P_1} + {P_2}\]

Step by step solution:
Note that the power of two given lenses are $P_1$ = +10D and $P_2$ = -5D.
Then the power of the combination of lenses is
\[
  P = {P_1} + {P_2} \\
  P = \left( { + 10 - 5} \right)D \\
  P = 5D \\
 \]
Calculate the focal length of the combination:
\[
  f = \dfrac{1}{P} \\
  f = \dfrac{1}{5}m \\
  f = \dfrac{{100}}{5}cm \\
  f = 20cm \\
\]
As the object has a virtual image and magnification 2, then you must take the magnification positive and write:
\[
  m = \dfrac{v}{u} \\
  \dfrac{v}{u} = 2 \\
  v = 2u \\
\]
Put the values of f and v in the lens formula to obtain the object distance.
\[
  \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} \\
  \dfrac{1}{{2u}} - \dfrac{1}{u} = \dfrac{1}{f} \\
  \dfrac{{1 - 2}}{{2u}} = \dfrac{1}{{20}} \\
  u = - 10 \\
\]
Hence the object must be placed at a distance of -10cm from the lens.
So option D. is the correct one.

Note: Remember the Cartesian sign conventions. The conventions are such that:
All distances must be measured from the optical center of the lens.
Distances measured along the direction of the incident ray that is along the +ve X axis are taken to be positive where the distances in the direction opposite to the direction of light that is along –ve X axis are taken to be negative.
Distances measured upward and perpendicular to the principal axis are taken to be positive and downward are taken to be negative.
When the image formed is real and when virtual. A positive magnification implies a virtual image where a negative magnification implies a real image.