In the arrangement shown in figure, image of light source O is obtained at the bottom of beaker containing water μ=43. The value of H is ___ (given that O is 36cm from pole of lens and mirror is at 100 cm from lens)

For lens  1v−1u=1f ⇒1v−1-36=130 ⇒v=180cm In absence of mirror, image should have formed at I1 , 80 cm behind mirror. So, I1 will behave as virtual object and its real image should have formed at I2, 80 cm vertically below principal axis. But, due to presence of water, real image is shifted to I3 , at bottom of beaker. Inside beaker, 2020-x=43 ⇒x=5 cm Thus, H=80+x=85 cm