Question

Question #424e1

Answer 1

`5"D"`

Explanation:


As image formed by the lens is real it follows that lens is convex.

Let object be located at a distance `u` form the lens. Image is given as formed on the other side at `v=20 cm`.

We know for that for a thin lens of focal length `f` the equation is

`1/v+1/u=1/f`

Keeping in view the sign convention the object distance is `-ve`. Measured from the center of lens on the other side (`-x` axis). If `f_1` is the focal length of this lens, inserting given values we get
`1/20−1/u=1/f_1` ..............(1)

When another lens is added to it, the image shifts `10 cm` towards the combination. This implies that second lens is also convex. Let second lens has focal length `f_2`

From lens' formula we know that focal length `F` of the combination is given by the equation
`1/F=1/f_1` `+1/f_2` .....(2)

Also from lens equation we get

`1/10−1/u=1/F` ......(3)

Now from equations (2) & (3), we get

`1/10−1/u=1/f_1` `+ 1/f_2` .....(4)
Subtracting (1) from (4)

`1/10−1/u-(1/20−1/u)=1/f_1` `+ 1/f_2` `-1/f_1`
`=>1/f_2` `=1/20`
`=>f_2` `=20cm=0.2m`

Thus Power P of second lens `=1/f_2` `=1/0.2=5"D"`