A converging lens of focal length 12 cm and a diverging mirror of focal length 7.5 cm are placed 5.0 cm apart with their principal axes coinciding. Where should an object be placed so that its image falls on itself?
Answers
Given :
f= 12cm
diverging mirror=- 7.5cm
let the object be placed at distance x from lens
for convex lens:
u= -x
f= -12cm
lens formula :
1/f= 1/v- 1/u
1/v=-1/12 -1/x
1/v= -x-12/ 12x
1/v= -(x+12)/12x
v= -[12x /x+12]
This image forms object for convex mirror
for mirror :
u= - (5+[12x/ x+12])
= -(17x+60/ x+12)
f= - 7. 5cm
From mirror formula :
1/f=1/v+1/u
1/v= - 1/7.5 + [ x+12/17x+60]
1/v= 17x+60 - 7.5/ 7.5 (17x+60)
v= 7.5 (17x+60)/ 52.5-127.5 x
v= 250(x+4)/15x-100
v= 50(x+4)/3x-20
For second refraction in concave lens :
u= -[ 5- 50(x+4)/ 3x-20]
v=+x
1/v-1/u=1/f
1/x+1/[ 5-50(x+4)/ 3x-20] = -1/20
25x²- 1400x-6000=0
x²-56x-240=0
(x-60)(x+4)=0
x= 60cm
object should be placed at a distance of 60cm from lens farther away from mirror so that final image is formed on itself
Answer:
ANSWER
Let the object be placed at a distance x cm away from the lens.
For convex lens, u=-x, f=-12 cm
From the lens formula,
v
1
−
u
1
=
f
1
v
1
=
−12
1
−
−x
1
v=−(
x+12
12x
)
Thus, the virtual image due to the 1st reflection lies in the same side as that of object.
u=−(5+
x+12
12x
)
=−(
x+12
17x+60
)
For the mirror of f=-7.5 cm, mirror equation is
v
1
+
u
1
=
f
1
v
1
=
−7.5
1
+
7x+60
x+12
v=
52.5−127.5x
7.5(17x+60)
v=
15x−100
250(x+4)
=
3x−20
50(x+4)
The image is formed towards the left of the mirror.
For the 2nd refraction in concave lens: u=−[
3x−20
5−50(x+4)
]
Assume the image is formed between the lens and mirror is 3x-20
v=x (Since the final image is formed on the object A''B'')
v
1
−
u
1
=
f
1
x
1
+
3x−20
[5−50(x+4)]
1
=
−20
1
=>25x
2
−1400x−6000=0
=>x
2
−5x−240=0
=>x=60,
The object should be placed at 60 cm from the lens and far away from the mirror. So the final image formed on itself