The focal lengths of the objective and the eyepiece are 1 cm and 4 cm respectively . The distance between them is 14.5 cm . If an object of height 1mm is placed at a distance of 1.1 cm from the objective what will be the position and the size of the image seen throught the microscope?

Explanation

Here , `f_o=1 cm and f_e =4 cm ` .
For the objective , u=-1.1cm. If the image of the object is formed at a distance v, we have from the equation of the lens,
`1/v+1/(1.1)=1/1 or, 1/v+(10)/(11)=1 or, v=11 cm`
So the image formed by the objective is at a distance 11 cm on the other side of the objective and this image real. This image acts as an object to the eyepiece.
`therefore m_1 =v/u=(11)/(1.1)=10`
Now , the object distance relative to the eyepiece =-(14.5-11)=-3.5 cm
If the image is formed at a distance of V from the eyepiece we have from the equation of the lens,
` 1/V+1/(3.5)=1/4 or, 1/V=-2/7+1/4=-1/(28) or, V=-28 cm `
This image is virtual and will be formed at a distance of 28 cm in front of the eyepiece.
`therefore m_2=(28)/(3.5)=8`
`therefore` Magnification of the final image ,
m=`m_1xxm_2=10xx8=80`
`therefore` The size of the final image `=80xx1=80mm =8 cm `