Question

The focal length of a thin biconvex lens is \[20cm\]. When an object is moved from a distance of \[25cm\] in front of it to \[50cm\], the magnification of its image changes from \[{m_{25}}\]​ to\[{m_{50}}\]​. The ratio of\[{m_{25}}\]​ to \[{m_{50}}\] is:
A) \[2\]
B) \[4\]
C) \[6\]
D) \[8\]

Answer 1

Hint: First we calculate distance of an image for given two distances of object as \[25cm\] and \[50cm\]. Then we calculate magnification by formula used in case of distance of an object and an image.

Formula used:
We are using lens formula \[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\] to calculate \[v\] in both distance of an object \[u\]. Magnification is calculated by \[m = \dfrac{{ - v}}{u}\].

Complete step by step solution:
Given: focal length of biconvex lens, \[f = 20\], distance of an object \[u = {\text{ }}25{\text{ }}cm\] and \[50\]\[cm\].
First we calculate the distance of the image for \[u = {\text{ }}25{\text{ }}cm\].
We know that the lens formula is given as
\[\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}\]
\[ \Rightarrow \dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\]
\[
  \dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{{( - 25)}} = \dfrac{1}{{20}} - \dfrac{1}{{25}} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{{5 - 4}}{{100}} = \dfrac{1}{{100}} \\
  v = 100{\text{ }}cm \\
 \]
Hence, the distance of the object for \[u = {\text{ }}25{\text{ }}cm\] is \[v = 100{\text{ }}cm\].
Magnification for \[u = {\text{ }}25{\text{ }}cm\] is given by, \[{m_{25}} = \dfrac{{ - v}}{u}\]
\[
  {m_{25}} = \dfrac{{ - v}}{u} \\
  {m_{25}} = \dfrac{{ - 100}}{{ - 25}} = 4 \\
 \]
Again we calculate distance of an image for
\[\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u}\]
\[
  \dfrac{1}{v} = \dfrac{1}{{20}} + \dfrac{1}{{( - 50)}} = \dfrac{1}{{20}} - \dfrac{1}{{50}} \\
   \Rightarrow \dfrac{1}{v} = \dfrac{{5 - 2}}{{100}} = \dfrac{3}{{100}} \\
  \therefore v = \dfrac{{100}}{3}{\text{ }}cm \\
 \]
Hence, distance of an image for \[u = - 50{\text{ }}cm\] is \[v = \dfrac{{100}}{3}{\text{ }}cm\].
And magnification for is given by, \[{m_{50}} = \dfrac{{ - v}}{u}\]
\[
  {m_{50}} = \dfrac{{ - \dfrac{{100}}{3}}}{{ - 50}} = \dfrac{{100}}{{50 \times 3}} \\
  {m_{50}} = \dfrac{2}{3} \\
 \]
Hence, the ratio of \[{m_{25}}\]to \[{m_{50}}\], i.e. \[{m_{25}}:{m_{50}}\] is given by
\[
  \dfrac{{{m_{25}}}}{{{m_{50}}}} = \dfrac{4}{{\dfrac{2}{3}}} \\
   \Rightarrow \dfrac{{{m_{25}}}}{{{m_{50}}}} = 4 \times \dfrac{3}{2} = 6 \\
  \therefore {m_{25}}:{m_{50}} = 6 \\
 \]
Hence, the required ratio is given by, \[{m_{25}}:{m_{50}} = 6\]

Therefore, the correct option is C.

Additional information:
The ratio of the height of an image to the height of an object is known as Magnification of a lens. It is also described as the ratio of image distance to the object distance. The distance of the object is \[u\], the distance of the image is \[v\]. Then we calculate magnification of a lens by the formula \[m = \dfrac{{ - v}}{u}\]. In case of height: height of image=\[h'\] and height of an object = \[h\]. Then, magnification of a lens is given by \[m = \dfrac{{h'}}{h}\].

Note: We know that distance taken in the lens may be positive or negative. Students must be careful to choose the sign for \[u\] and \[v\]. Students must choose negative signs for distance taken for objects, i.e., \[u\].
Here magnification is calculated by only with \[m = \dfrac{{ - v}}{u}\] not by \[m = \dfrac{{h'}}{h}\]. We can’t use \[m = \dfrac{{h'}}{h}\] because in question the height of the image and object is not given. Students must choose the correct formula for calculating magnification.