CBSE Class 12-science Answered
A concave mirror produces a real image and the virtual image of the same magnification at distance of 0.12 m and 0.08 m respectively. The radius of curvature of mirror is
Asked by viharikashankar | 18 Jun, 2019, 22:42: PM
Virtual image is produced at 0.08 m (behind the mirror ) and real image is produced at 0.12 m (in front of mirror).
Let u1 be the object-to-mirror distance for virtual image and u2 be the object-to-mirror distance for real image.
since magnification is same for both the cases, we have, .08/u1 = 0.12/u2 or ( u1 / u2 ) = 2/3 ...................(1)
Let f be the focal length of concave mirror
mirror equation for virtual image :-
.................... (2)
![begin mathsize 14px style fraction numerator negative 1 over denominator 0.08 end fraction space plus space 1 over u subscript 1 space equals space 1 over f end style](https://images.topperlearning.com/topper/tinymce/cache/dc8cf8b0f4275a17d4462ff833ba680b.png)
mirror equation for real image :-
........................(3)
![begin mathsize 14px style fraction numerator 1 over denominator 0.12 end fraction space plus space 1 over u subscript 2 space equals fraction numerator begin display style 1 end style over denominator begin display style 0.12 end style end fraction space plus space fraction numerator begin display style bevelled 2 over 3 end style over denominator begin display style u subscript 1 end style end fraction space equals space 1 over f space end style](https://images.topperlearning.com/topper/tinymce/cache/a492538fed4af116d04a4b33aa0365da.png)
In eqn.(3), we substituted for u2 from eqn.(1)
Since RHS of eqn.(2) and (3) are same, we equate both LHS of eqn.(2) and (3) and solve the formed equation to get u1
![begin mathsize 14px style fraction numerator negative 1 over denominator 0.08 end fraction plus 1 over u subscript 1 space equals space fraction numerator 1 over denominator 0.12 end fraction plus fraction numerator begin display style bevelled 2 over 3 end style over denominator u subscript 1 end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/624ed012b6fb9bbf47044fb1ccded8fc.png)
Hence u2 = (3/2)u1 = (3/2)×0.016 = 0.024 m
By substituting u1 or u2 in eqn.(2) or (3), we get f = 0.02 m.
Hence radius of curvature of mirror = 0.04 m
Answered by Thiyagarajan K | 19 Jun, 2019, 16:37: PM
Concept Videos
CBSE 12-science - Physics
Asked by jothisugashini216 | 16 Jul, 2024, 20:30: PM
CBSE 12-science - Physics
Asked by prithviraj.chopra2011 | 14 Jul, 2024, 23:00: PM
CBSE 12-science - Physics
Asked by basithhhabduuu | 14 Jul, 2024, 17:07: PM
CBSE 12-science - Physics
Asked by asmaabid356 | 10 Jul, 2024, 21:03: PM
CBSE 12-science - Physics
Asked by pushpamagadum21 | 09 Jul, 2024, 20:43: PM
CBSE 12-science - Physics
Asked by sumitghorband09 | 20 Jun, 2024, 21:42: PM
CBSE 12-science - Physics
Asked by axonuploadserver4 | 14 Jun, 2024, 17:41: PM
CBSE 12-science - Physics
Asked by trishaktiprasadmallik693 | 11 Jun, 2024, 07:17: AM
CBSE 12-science - Physics
Asked by rahuldevgayen4 | 10 Jun, 2024, 14:58: PM