An equi-convex lens of `mu=1.5 and R=20 cm` is cut into two equal parts along its axis. Two parts are then separated by a distance of `120 cm` (as shown in figure). An object of height `3 mm` is placed at a distance of `30 cm` to the left of first half lens. The final image will form at

Options

A. `120 cm` to the right of first half lens, `3 mm` in size and inverted
B. `150 cm` to the right of first half lens, `3 mm` in size and erect
C. `120 cm` to the right of first half lens, `4 mm` in size and inverted
D. `150 cm` to the right of first half lens, `4 mm` in size and erect

Answer

B

Explanation

`1/f=(1.5-1)(1/20-1/(-20))`
`rArr f=+20 cm`
`1/v_1-1/(-30)=1/(+20)`
`:. v_1=+60 cm`
`m_1=(v_1)/(u_1)=(+60)/(-30)=-2`
For second lens,
`1/(v_2)-1/(-60)=1/(+20)`
`v_2=+30 cm`
`m_2=v_2/u_2=(+30)/(-60)=-1/2`
`m=m_1m_2=1`
Final image is `30 cm` to the right of second lens or
`150 cm` to the right of first lens.
`m=1`
Hence, image height =object height
`=3 mm`