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JEE Main Physics Class 12 Ray Optics and Optical Instruments Part 1 Questions
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© examsnet.com
Question : 74
Total: 99
An object at 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?
[2012]
7.2 m
2.4 m
3.2 m
5.6 m
Validate
Solution:
The focal length of the lens
1
f
=
1
v
−
1
u
=
1
12
+
1
240
=
20
+
1
240
=
21
240
f
=
240
21
c
m
When glass plate is interposed between lens and film, so shift produced will be
Shift
=
t
(
1
−
1
µ
)
1
(
1
−
1
3
∕
2
)
=
1
×
1
3
Now image should be form at
v
′
=
12
−
1
3
=
35
3
c
m
Now the object distance u.
Using lens formula again
1
f
=
1
v
′
−
1
u
⇒
1
u
=
1
v
′
−
1
f
⇒
1
u
=
3
35
−
21
240
=
1
5
[
3
7
−
21
48
]
⇒
1
u
=
1
5
[
48
−
49
7
×
16
]
⇒
u
=
−
7
×
16
×
5
=
−
560
c
m
=
−
5.6
m
© examsnet.com
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