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<

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