Answer:
Explanation:
Focal length of diverging lens,
fd= -33.5 cm
focal length of converging lens,
fc = 20 cm
Now let s is object distance and s' is image distance.
Now using lens formula for conversing lens
\frac1{f_c}=\frac1{s}+\frac1{s'}
f
c
1
=
s
1
+
s
′
1
Putting all values(object is at s=infinity)
Then image distance
\frac1{s'}= \frac1{f_c}- \frac1{s}
s
′
1
=
f
c
1
−
s
1
=\frac1{20}- \frac1{\infin}=
20
1
−
∞
1
s'=20cms
′
=20cm
Now using lens formula for diversing lens
s''=14-20=-6cm
Using lens formula
\frac{1}{-33.5} = \frac{1}{-6 }+ \frac{1}{s''' }\\s'''=\frac{(6)\times(-33.5)}{(-33.5+6)}\\s'''=\frac{(-201) }{(-27.5) }
−33.5
1
=
−6
1
+
s
′′′
1
s
′′′
=
(−33.5+6)
(6)×(−33.5)
s
′′′
=
(−27.5)
(−201)
image distance for diverging lens
s''' = 7.3 cm (behind the diverging lens)
