Let us solve it using the following lens maker formula
color(blue)(1/f=(mu-1)(1/R_1-1/R_2)........[1])
Where
f->"focal length of a lens in air"
mu->"refractive index of the lens w.r to air"
R_1->"radius of curvature of the surface of lens"
" facing incident rays"
R_2->"radius of curvature of the surface of lens"
" leaving emergent rays"
Calculation of focal length of first lens f_1
Here mu=mu_1;R_1=+R ;R_2=-R;f=f_1
From equation [1]
color(green)(1/(f_1) =(mu_1-1)(1/R-1/(-R))=(2(mu_1-1))/R
Calculation of focal length of second plano concave lens f_2
Here mu=mu_2;R_1=-R ;R_2=oo;f=f_2
From equation [1]
color(red)(1/(f_2) =(mu_2-1)(1/(-R)-1/(oo))=(-(mu_2-1))/R
If F denotes the equivalent focal length of the combination then
1/F=1/(f_1) +1/f_2
=>1/F=1/R(2mu_1 -2-mu_2 +1)
=>1/F=1/R(2mu_1 -mu_2 -1)
=>color(red)(F=R/(2mu_1 -(mu_2 +1))......[2])
a) So this expression represents the equivalent focal length of the combination.
b) To obtain the condition when the combination acts as diverging lens in air,we can say that the focus of the combination will be virtual one and by sign convention it will be negative.
Hence F<0
=>R/(2mu_1 -(mu_2 +1))<0
=>2mu_1 -(mu_2 +1)<0
=>mu_1 <(mu_2 +1)/2->"required condition"
c)From equation [2] we get
color(red)(F=R/(2mu_1 -(mu_2 +1))
=>color(red)(F=R/(2(mu_1 -(mu_2 +1)/2))
This equation suggests that for the given condition
mu_1 >(mu_2 +1)/2
the F will be positive and the combination will act as converging lens. When the object is kept far away from the combination it will form a real image in very diminished size at its focal plane. The ray diagram can then be drawn easily from this view point and it will be like this.
![drawn]()
