Consider a concave mirror and a convex lens (refractive$$ $$index=$$ $$1.5) of focal length $$10cm$$ each, separated by a distance of $$50cm$$ in air (refractive$$ index $$=$$ $$1) as shown in the figure. An object is placed at a distance of $$15$$ cm from the mirror. Its erect image formed by this combination has magnification $$M1$$. When the $$set-up$$ is kept in a medium of refractive index $$7/6$$, the magnification becomes $$M2$$. The magnitude $$M2M1$$ is

Question

Consider a concave mirror and a convex lens (refractive$$ $$index=$$ $$1.5) of focal length $$10cm$$ each, separated by a distance of $$50cm$$ in air (refractive$$ index $$=$$ $$1) as shown in the figure. An object is placed at a distance of $$15$$ cm from the mirror. Its erect image formed by this combination has magnification  $$M1$$.  When the $$set-up$$ is kept in a medium of refractive index $$7/6$$, the magnification becomes $$M2$$. The magnitude $$M2M1$$  is

Infinity Learn · Accepted Answer

$$1st$$ case:Reflection from $$mirror1f=1v+1u$$​⇒$$1$$−$$10=1v+1$$−$$15$$​⇒$$v=$$−$$30$$ ​For lens​$$110=1v$$−$$1$$−$$20$$​⇒$$v=20$$​$$M1=m1$$×$$m2=v1u1$$×$$v2u2=$$−$$30$$−$$15$$×$$20$$−$$20=2$$×$$1=2$$  in $$air2nd$$ case :For $$mirrorv=$$−$$30$$​For lens​$$1v$$−$$1$$−$$20=3/27/6$$−$$12R$$⇒$$1v+120=27210=470$$∵$$1fair=$$μ−$$12R$$⇒$$110=1.5$$−$$12R$$⇒$$R=10cm1v=470$$−$$120$$​⇒$$v=140$$​$$M2=m1m2=2$$×$$140$$−$$20=$$−$$14$$​​Hence, $$M2M1=142=7$$

Best Courses for You

Detailed Solution

courses

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

download the app

Download the App