Question

Find the position of the image formed by the lens combination given in the figure.

Answer 1

The figure having two lenses and the focal length with the sign.
We should be aware of the combined effect formula to solve this question.
We have to use this effect for different types of length.

Complete step-by-step solution:
Let us consider.
the focal length that is denoted by $f$
the object’s distance is denoted by $u$
the image distance is $v$
From the lense
The lense formula is used to find the image position as follows,
${f_2} = - 10cm$
${u_2} = (15 - 5)cm$
Now ${u_2}$is,
${u_2} = 10cm$
$\dfrac{1}{{{v_3}}} = \dfrac{1}{{30}} + \dfrac{1}{\infty }$
From this ${v_3}$is,
${v_3} = 30cm$
And ${f_3} = 30cm$
${f_2} = - 10cm$
${u_2} = (15 - 5)cm$
${u_2} = 10cm$
$\dfrac{1}{{{v_3}}} = \dfrac{1}{{30}} + \dfrac{1}{\infty }$
${v_3} = 30cm$
${f_3} = 30cm$ ${f_1} = + 10cm$
${u_1} = - 30cm$
By using the lens formula,
$\dfrac{1}{{{v_1}}} - \dfrac{1}{{{u_1}}} = \dfrac{1}{{{f_1}}}$
Now rearranging the lens formula it becomes,
$\dfrac{1}{{{v_1}}} = \dfrac{1}{{{u_1}}} + \dfrac{1}{{{f_1}}}$
By using the lens formula ${v_1} = 15cm$
The first lense is formed at a distance $15cm$ and hence the image is formed.
The lens acts as a source of a second lens and its place at the right side of the first lense
In the case of the second lens,
${f_2} = - 10cm$
${u_2} = (15 - 5)cm$
Now ${u_2}$becomes,
${u_2} = 10cm$
After substituting the values the equation becomes,
$\dfrac{1}{{{v_2}}} = \dfrac{1}{{{u_2}}} + \dfrac{1}{{{f_2}}}$
Now substitute the value,
$\dfrac{1}{{{v_2}}} = - \dfrac{1}{{10}} + \dfrac{1}{{10}}$
After solving the above equation,
${v_2} = \infty$
The real image is formed at the second lens at an infinite distance. The image acts as an object for the lens which is placed in front of the current image in the third lens
${f_3} = 30cm$
${u_3} = \infty$
Now,
$\dfrac{1}{{{v_3}}} = \dfrac{1}{{{f_3}}} + \dfrac{1}{{{u_3}}}$
Now the equation becomes,
$\dfrac{1}{{{v_3}}} = \dfrac{1}{{30}} + \dfrac{1}{\infty }$
Hence, ${v_3} = 30cm$
So, the image is formed $30cm$to the right side of the third lens.

Note: We should understand about the lens used in combination
We should know about the lens formula used.
And should know the direction to be considered as positive and how to proceed with it.