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Two similar thin equi-convex lenses, of focal length f f f each, are kept coaxially in contact with each other such that the focal length of the combination is F 1 . F_1. F 1 . When the space between the two lenses is filled with glycerin ( μ = 1.5 ) (\mu = 1.5) ( μ = 1.5 ) , the equivalent focal length is F 2 . F_2. F 2 . The ratio F 1 : F 2 F_1 : F_2 F 1 : F 2 will be:
hard
Ray Optics and Optical Instruments
2019
physics
Explanation To solve this problem, we need to understand the behavior of the lens combination. \\
Given: \\
• Two similar thin equi-convex lenses with focal length f f f each. \\
• Focal length of the combination without glycerin is F 1 . F_1. \\
F 1 .
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• Equivalent focal length when the space is filled with glycerin is
Step 1: Calculate
When two lenses are in contact, the equivalent focal length
is given by:
1 F 1 = 1 f + 1 f 1 F 1 = 2 f F 1 = f 2 \\
\frac{1}{F_1} = \frac{1}{f} + \frac{1}{f} \\
\frac{1}{F_1} = \frac{2}{f} \\
F_1 = \frac{f}{2} \\
F 1 1 = f 1 + f 1 F 1 1 = f 2 F 1 = 2 f Step 2: Calculate
when glycerin is filled between the lenses
The focal length of the combination when a medium of refractive index
is filled between the lenses is given by:
1 F 2 = 1 f + 1 f − ( μ − 1 ) ⋅ d μ ⋅ f 2 \\
\frac{1}{F_2} = \frac{1}{f} + \frac{1}{f} - \frac{(\mu - 1) \cdot d}{\mu \cdot f^2} \\
F 2 1 = f 1 + f 1 − μ ⋅ f 2 ( μ − 1 ) ⋅ d Since the lenses are in contact, the separation
Thus, the term involving
becomes zero:
1 F 2 = 1 f + 1 f 1 F 2 = 2 f F 2 = f 2 \\
\frac{1}{F_2} = \frac{1}{f} + \frac{1}{f} \\
\frac{1}{F_2} = \frac{2}{f} \\
F_2 = \frac{f}{2} \\
F 2 1 = f 1 + f 1 F 2 1 = f 2 F 2 = 2 f Step 3: Calculate the ratio
F 1 : F 2 F_1 : F_2 \\
F 1 : F 2 Since both
and
are equal to
the ratio is:
F 1 : F 2 = f 2 : f 2 = 1 : 1 \\
F_1 : F_2 = \frac{f}{2} : \frac{f}{2} = 1 : 1 \\
F 1 : F 2 = 2 f : 2 f = 1 : 1 However, the question asks for the ratio when glycerin is filled, which implies a different interpretation.
Re-evaluate the scenario considering the refractive index change:
The correct interpretation should consider the effective focal length change due to glycerin.
Recalculate
with the correct interpretation:
When glycerin is filled, the effective focal length changes due to the refractive index:
1 F 2 = 1 f + 1 f − ( μ − 1 ) f 1 F 2 = 2 f − 0.5 f 1 F 2 = 1.5 f F 2 = f 1.5 = 2 f 3 \\
\frac{1}{F_2} = \frac{1}{f} + \frac{1}{f} - \frac{(\mu - 1)}{f} \\
\frac{1}{F_2} = \frac{2}{f} - \frac{0.5}{f} \\
\frac{1}{F_2} = \frac{1.5}{f} \\
F_2 = \frac{f}{1.5} = \frac{2f}{3} \\
F 2 1 = f 1 + f 1 − f ( μ − 1 ) F 2 1 = f 2 − f 0.5 F 2 1 = f 1.5 F 2 = 1.5 f = 3 2 f Now, calculate the ratio
F 1 : F 2 : F 1 : F 2 = f 2 : 2 f 3 = 3 6 : 4 6 = 3 : 4 F_1 : F_2: \\
F_1 : F_2 = \frac{f}{2} : \frac{2f}{3} \\
= \frac{3}{6} : \frac{4}{6} \\
= 3 : 4 \\
F 1 : F 2 : F 1 : F 2 = 2 f : 3 2 f = 6 3 : 6 4 = 3 : 4 Therefore, the correct ratio is
This corresponds to Option 4.