(i) Magnification for the system is \[+1\] |
(ii) Magnification for the system is \[-1\] |
(iii) Final image by the system will be real and at distance of \[110\text{ }cm\]from centre of curvature of spherical mirror |
(iv) Final image by the system will be real and at distance of \[60\text{ }cm\]from centre of curvature of spherical mirror |
The correct statement is |
A) (i) and (iv)
B) (ii) and (iii)
C) (i) and (iii)
D) (ii) and (iv)
Correct Answer: B
Solution :
[b] For the 1st refraction through the lens \[\frac{1}{v}+\frac{1}{40}=\frac{1}{20}\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{40}=\frac{1}{40}\Rightarrow v=40cm\] Distance of this image from the mirror is \[d=7040=30,\]i.e. the centre of curvature of the mirror, therefore ray retraces its path and image coincides with the object, and it will be inverted. Hence final image will be at distance \[40\text{ }cm+70\text{ }cm=110\text{ }cm\]from the mirror.You need to login to perform this action.
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