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Long Answer Questions · Q7

Q.Obtain the lens maker's formula and mention its significance.

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Step 1. A thin lens of refractive index n2n_2 sits in a medium of index n1n_1, with surfaces of radii R1,R2R_1,R_2. For a point object OO, refraction at the first surface (from n1n_1 to n2n_2) gives an intermediate image at v′v': n2v′−n1u=n2−n1R1\dfrac{n_2}{v'}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R_1}.

Step 2. This intermediate image acts as the object for the second surface (from n2n_2 back to n1n_1), forming the final image at vv: n1v−n2v′=n1−n2R2\dfrac{n_1}{v}-\dfrac{n_2}{v'}=\dfrac{n_1-n_2}{R_2}.

Step 3. Adding the two equations cancels the intermediate n2/v′n_2/v' terms: n1v−n1u=(n2−n1)(1R1−1R2)\dfrac{n_1}{v}-\dfrac{n_1}{u}=(n_2-n_1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right), i.e. 1v−1u=(n2n1−1)(1R1−1R2)\dfrac{1}{v}-\dfrac{1}{u}=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right).

Step 4. For an object at infinity (u→∞u\to\infty), the image forms at the focus (v→fv\to f), so 1f=(n2n1−1)(1R1−1R2)\dfrac{1}{f}=\left(\dfrac{n_2}{n_1}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right); for a lens of index nn in air (n1=1,n2=nn_1=1,n_2=n), this is 1f=(n−1)(1R1−1R2)\boxed{\dfrac{1}{f}=(n-1)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)}, the lens maker's formula. …

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