Question

When a thin convex lens is immersed in a liquid of refractive index $1.2$, the focal length of the lens becomes $48\text{ cm}$. If the ratio of the radii of curvature of the lens is $2:3$ and the refractive index of the material of the lens is $1.5$, then the radii of curvature of the lens are:

• A. $20\text{ cm},\,30\text{ cm}$
• B. $10\text{ cm},\,15\text{ cm}$
• C. $08\text{ cm},\,12\text{ cm}$
• D. $16\text{ cm},\,24\text{ cm}$

Hint:

Whenever a lens is immersed in a liquid, do not use the Lens Maker's Formula for air. Always replace the refractive index term by \[ \left(\frac{\mu_g}{\mu_m}-1\right). \] For a convex lens, \[ R_1>0,\qquad R_2<0, \] which often converts \[ \left(\frac{1}{R_1}-\frac{1}{R_2}\right) \] into the sum of the reciprocals of the magnitudes of the radii.

Answer 1

The Correct Option is A

Concept: When a lens is placed in a medium other than air, its focal length changes because the refractive index contrast between the lens material and the surrounding medium changes. The appropriate relation in such cases is the Lens Maker's Formula in a medium: \[ \frac{1}{f} = \left( \frac{\mu_g}{\mu_m}-1 \right) \left( \frac{1}{R_1}-\frac{1}{R_2} \right), \] where

• $f$ = focal length of the lens in the medium,

• $\mu_g$ = refractive index of the lens material,

• $\mu_m$ = refractive index of the surrounding medium,

• $R_1$ and $R_2$ = radii of curvature of the two lens surfaces.
For a convex lens, the first surface has a positive radius of curvature and the second surface has a negative radius of curvature according to the Cartesian sign convention. The problem provides the focal length of the lens inside a liquid and the ratio of the radii of curvature. Therefore, we first express the radii in terms of a common variable and then apply the Lens Maker's Formula to determine their actual values.

Step 1: Write all the given quantities and assign variables to the radii of curvature. From the question, \[ \mu_g=1.5 \] \[ \mu_m=1.2 \] \[ f=48\text{ cm} \] Also, \[ R_1:R_2=2:3. \] Let \[ R_1=2x \] and \[ R_2=3x. \] Since the lens is convex, \[ R_1=+2x \] and \[ R_2=-3x. \] These sign conventions are extremely important because the Lens Maker's Formula involves the quantity \[ \left(\frac{1}{R_1}-\frac{1}{R_2}\right). \]

Step 2: Substitute the refractive indices into the Lens Maker's Formula. Using \[ \frac{1}{f} = \left( \frac{\mu_g}{\mu_m}-1 \right) \left( \frac{1}{R_1}-\frac{1}{R_2} \right), \] we get \[ \frac{1}{48} = \left( \frac{1.5}{1.2}-1 \right) \left( \frac{1}{2x}-\frac{1}{-3x} \right). \] Now simplify the refractive-index term: \[ \frac{1.5}{1.2} = \frac{15}{12} = \frac{5}{4}. \] Therefore, \[ \frac{5}{4}-1 = \frac{1}{4}. \] Hence, \[ \frac{1}{48} = \frac{1}{4} \left( \frac{1}{2x}+\frac{1}{3x} \right). \]

Step 3: Simplify the curvature term carefully. Consider \[ \frac{1}{2x}+\frac{1}{3x}. \] Taking LCM, \[ \frac{1}{2x}+\frac{1}{3x} = \frac{3+2}{6x} = \frac{5}{6x}. \] Substituting into the equation, \[ \frac{1}{48} = \frac{1}{4}\times\frac{5}{6x}. \] Therefore, \[ \frac{1}{48} = \frac{5}{24x}. \] This equation contains only one unknown quantity, namely $x$.

Step 4: Solve for the value of $x$. Cross-multiplying, \[ 24x=48\times5. \] Thus, \[ 24x=240. \] Dividing both sides by $24$, \[ x=\frac{240}{24}. \] \[ x=10. \] Therefore, \[ x=10\text{ cm}. \]

Step 5: Calculate the actual radii of curvature. Since \[ R_1=2x, \] we obtain \[ R_1=2(10)=20\text{ cm}. \] Similarly, \[ R_2=3x, \] which gives \[ R_2=3(10)=30\text{ cm}. \] Thus, the magnitudes of the radii of curvature are \[ 20\text{ cm} \] and \[ 30\text{ cm}. \]

Final Conclusion: After applying the Lens Maker's Formula in a liquid medium and using the given ratio of radii of curvature, we obtain \[ R_1=20\text{ cm} \] and \[ R_2=30\text{ cm}. \] Hence, the radii of curvature of the lens are \[ \boxed{20\text{ cm},\,30\text{ cm}}. \] Therefore, option (A) is the correct answer.