A thin convex lens is made of two materials with refractive indices $$n1$$ and $$n2$$, as shown in figure. The radius of curvature of the left and right spherical surface are equal. f is the focal length of the lens when $$n1=n2=n$$. The focal length is $$f+$$Δf when $$n1=n$$ and $$n2=n+$$Δn . Assuming Δn

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A thin convex lens is made of two materials with refractive indices $$n1$$ and $$n2$$, as shown in figure. The radius of curvature of the left and right spherical surface are equal.  f is the focal length of the lens when  $$n1=n2=n$$. The focal length is $$f+$$Δf  when $$n1=n$$ and $$n2=n+$$Δn . Assuming Δn<

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$$f=R2n$$−$$1____1In$$ $$2nd$$ case,$$1f+df=n$$−$$1R+n+dn$$−$$1R$$ ⇒$$1f+df=1f+dnR$$ Clearly, if dn<$$0$$, R.H.S value is less $$than1f$$ ∴df>$$0i$$.e (1) is $$correctOption(2)f=R21.5$$−$$1$$⇒$$R=20cm$$ Now, $$1f+df=1f+dnR$$⇒$$120+df=120+dn20$$⇒$$1201+df20=120(1+dn)$$⇒$$1$$−$$df20=1+dn$$⇒$$df=$$−$$20dn$$ ∴$$df=0.02cm$$ (2) is $$correctOption(3)$$: If convex surfaces are replaced with concave surfaces, sign of R will $$change1f1+dff=1f+dnR=1f+dn2fn$$−$$1Since$$, $$f=R2n$$−$$1$$⇒$$1R=12fn$$−$$1$$ ∴$$1$$−$$dfff=1f1+dn2n1$$−$$1n$$⇒−$$dff=dnn121$$−$$1n$$ As this is independent of R, (3) is correct. Option (4): $$dff=dnn121$$−$$1n$$ For the range $$1$$ < n < $$2$$, $$21$$−$$1n$$ is always less than $$1$$ i.e, dff>dnn (4) is wrong

A thin convex lens is made of two materials with refractive indices n1 and n2, as shown in figure. The radius of curvature of the left and right spherical surface are equal. f is the focal length of the lens when n1=n2=n. The focal length is f+Δf when n1=n and n2=n+Δn . Assuming Δn<

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