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A thin convex lens of refractive index 1.5 has 20 cm focal length in air. If the lens is completely immersed in a liquid of refractive index 1.6, us focal length will be
- a)– 160 cm,
- b)– 100 cm
- c)+ 10 cm
- d)+ 100 cm
Correct answer is option 'A'. Can you explain this answer?
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A thin convex lens of refractive index 1.5 has 20 cm focal length in a...
Using the formula $\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, where $f$ is the focal length, $n$ is the refractive index, and $R_1$ and $R_2$ are the radii of curvature of the lens surfaces, we can solve for the new focal length $f'$:
In air:
$n_1=1$ (air)
$n_2=1.5$ (lens)
$f=20\text{ cm}$
$\frac{1}{20}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Simplifying:
$R_1=R_2=10\text{ cm}$ (since it's a thin lens, the radii of curvature are the same and equal to half the focal length)
When the lens is immersed in a liquid of refractive index 1.6, we have:
$n_1=1.6$ (liquid)
$n_2=1.5$ (lens)
$f'=?$
$\frac{1}{f'}=(1.5-1.6)\left(\frac{1}{10}-\frac{1}{10}\right)$
Simplifying:
$\frac{1}{f'}=0.1\times0$
Since the term on the right-hand side is zero, we can't solve for $f'$ using this formula. However, we know that when a lens is immersed in a medium of higher refractive index, its focal length decreases. Therefore, we can conclude that the focal length of the lens will be less than 20 cm when it is immersed in a liquid of refractive index 1.6.
In air:
$n_1=1$ (air)
$n_2=1.5$ (lens)
$f=20\text{ cm}$
$\frac{1}{20}=(1.5-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
Simplifying:
$R_1=R_2=10\text{ cm}$ (since it's a thin lens, the radii of curvature are the same and equal to half the focal length)
When the lens is immersed in a liquid of refractive index 1.6, we have:
$n_1=1.6$ (liquid)
$n_2=1.5$ (lens)
$f'=?$
$\frac{1}{f'}=(1.5-1.6)\left(\frac{1}{10}-\frac{1}{10}\right)$
Simplifying:
$\frac{1}{f'}=0.1\times0$
Since the term on the right-hand side is zero, we can't solve for $f'$ using this formula. However, we know that when a lens is immersed in a medium of higher refractive index, its focal length decreases. Therefore, we can conclude that the focal length of the lens will be less than 20 cm when it is immersed in a liquid of refractive index 1.6.
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A thin convex lens of refractive index 1.5 has 20 cm focal length in air. If the lens is completely immersed in a liquid of refractive index 1.6, us focal length will bea)– 160 cm,b)– 100 cmc)+ 10 cmd)+ 100 cmCorrect answer is option 'A'. Can you explain this answer?
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