Question

Please help for physics question?

Answer 1

I got option (C)

Explanation:

As described in the problem the given optical system may be treated as a combination of three lenses where there are two equi-convex glass lenses of focal lengths `f_1=10cm and f_2=20cm` in contact and one concave water lens formed by inserted water in between the gap of two glass lenses.

Now let the radius of curvature of the equi-convex lens of focal length `f_1` be `R_1`. So by using lens maker formula we can write

`1/(f_1) = (mu_g-1)(1/R_1+1/R_1)`

`=>1/10 = (3/2-1)*2/R_1`

`=R_1=10cm`

Similarly if the radius of curvature of the equi-convex lens of focal length `f_2` be `R_2`. then by using lens maker formula we can write

`1/(f_2) = (mu_g-1)(1/R_2+1/R_2)`

`=>1/20 = (3/2-1)*2/R_2`

`=R_2=20cm`

So the water lens will have two concave surfaces with radius of curvature `R_1andR_2`

And the focal length of water lens `f_w` will be given by

`1/(f_w) = (mu_w-1)(-1/R_1-1/R_2)`

`=>1/(f_w) = (4/3-1)(-1/10-1/20)`

`=>1/(f_w) = -1/3*3/20=-1/20`

So `f_w=-20cm`

Now if the focal length of the combination of these three lenses be `f_c` then

`1/(f_c)=1/(f_1)+1/(f_w)+1/(f_2)`

`=>1/(f_c)=1/10-1/20+1/20`

`=>f_c=10cm`

Let us consider that the combination forms real image of twice in size of an object when object distance is `u_1`. In this case its image distance will be `-2u_1`,as magnification is `-2` for real image here.

So by lens formula we have

`1/(-2u_1)-1/(u_1)=1/(f_c)`

`=>-(1+2)/(2u_1)=1/10`

`=>u_1=-15cm` (here -ve sign denotes the real object distance)

Again if the combination forms virtual image of twice in size of an object when object distance is `u_2`, then its image distance will be `2u_2`,as magnification is `+2` for virtual image here.

So by lens formula we have

`1/(2u_2)-1/(u_2)=1/(f_c)`

`=>(1-2)/(2u_2)=1/10`

`=>u_2=-5cm` (here -ve sign denotes the real object distance)

Important Note proposed by renowned and respected teacher known to us as `color(magenta)"A08"`.

"Two shortcuts.

First for equi-convex lens R=f.

Second when a convex of focal length 20cm and a concave lens of focal length -20cm are placed side by side the focal length of combination is ∞. Hence, equivalent focal length of combination of three lenses reduces to the focal length of first lens."