Two convex lenses of focal lengths f1 and f2 form images with magnification m1 and m2, when used individually for an object kept at the same distance from the lenses. Then, f1 / f2 is

Question

Two convex lenses of focal lengths f1 and f2 form images with magnification m1 and m2, when used individually for an object kept at the same distance from the lenses. Then, f1 / f2 is (A) (m1(1-m1)/m2(1-m2)) (B) (m1(1-m2)/m2(1-m1)) (C) (m2(1-m1)/m1(1-m2)) (D) (m2(1-m2)/m1(1-m1))

Tardigrade Admin · Accepted Answer

The linear magnification of lens in terms of focal length f and u. m=(f/f+u) For first lens, u1=((1-m1/m1)) t1 For second lens, u2=((1-m2/m2)) f2 According to the question, u1 =u2 ∴ ((m1-1)/m1) f1 =((m2-1)/m2) f2 (f1/f2) =(m1/m2) ⋅ ((m2-1)/(m1-1)) =(m1(m2-1)/m2(m1-1))

Q. Two convex lenses of focal lengths $f_{1}$ and $f_{2}$ form images with magnification $m_{1}$ and $m_{2}$ , when used individually for an object kept at the same distance from the lenses. Then, $f_{1} / f_{2}$ is

$\frac{m _{1} ( 1 - m _{1} )}{m _{2} ( 1 - m _{2} )}$

$\frac{m _{1} ( 1 - m _{2} )}{m _{2} ( 1 - m _{1} )}$

$\frac{m _{2} ( 1 - m _{1} )}{m _{1} ( 1 - m _{2} )}$

$\frac{m _{2} ( 1 - m _{2} )}{m _{1} ( 1 - m _{1} )}$

Q. Two convex lenses of focal lengths f1​ and f2​ form images with magnification m1​ and m2​, when used individually for an object kept at the same distance from the lenses. Then, f1​/f2​ is

m2​(1−m2​)m1​(1−m1​)​

m2​(1−m1​)m1​(1−m2​)​

m1​(1−m2​)m2​(1−m1​)​

m1​(1−m1​)m2​(1−m2​)​

Q. Two convex lenses of focal lengths $f_{1}$ and $f_{2}$ form images with magnification $m_{1}$ and $m_{2}$ , when used individually for an object kept at the same distance from the lenses. Then, $f_{1} / f_{2}$ is

$\frac{m _{1} ( 1 - m _{1} )}{m _{2} ( 1 - m _{2} )}$

$\frac{m _{1} ( 1 - m _{2} )}{m _{2} ( 1 - m _{1} )}$

$\frac{m _{2} ( 1 - m _{1} )}{m _{1} ( 1 - m _{2} )}$

$\frac{m _{2} ( 1 - m _{2} )}{m _{1} ( 1 - m _{1} )}$