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Two identical thin plano-convex glass lenses (refractive index 1.5 ) 1.5) 1.5 ) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. 1.7. 1.7. The focal length of the combination is
hard
Ray Optics and Optical Instruments
2015
physics
Explanation To solve this problem, we need to find the focal length of the lens combination. \\
Given: \\
• Refractive index of the glass lenses ( n 1 ) = 1.5 (n_1) = 1.5 \\
( n 1 ) = 1.5 • Refractive index of the oil ( n 2 ) = 1.7 (n_2) = 1.7 \\
( n 2 ) = 1.7
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• Radius of curvature of each lens
cm
Each plano-convex lens has one flat side and one convex side.
The focal length of a single plano-convex lens is given by the lens maker's formula:
1 f 1 = ( n 1 − 1 ) ( 1 R 1 − 1 R 2 ) \\
\frac{1}{f_1} = (n_1 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) \\
f 1 1 = ( n 1 − 1 ) ( R 1 1 − R 2 1 ) For a plano-convex lens:
R 1 = R \\
R_1 = R \quad R 1 = R and
R 2 = ∞ \quad R_2 = \infty \\
R 2 = ∞ Thus, the formula becomes:
1 f 1 = ( 1.5 − 1 ) ( 1 20 − 1 ∞ ) 1 f 1 = 0.5 × 1 20 1 f 1 = 0.5 20 f 1 = 40 \\
\frac{1}{f_1} = (1.5 - 1) \left(\frac{1}{20} - \frac{1}{\infty}\right) \\
\frac{1}{f_1} = 0.5 \times \frac{1}{20} \\
\frac{1}{f_1} = \frac{0.5}{20} \\
f_1 = 40 f 1 1 = ( 1.5 − 1 ) ( 20 1 − ∞ 1 ) f 1 1 = 0.5 × 20 1 f 1 1 = 20 0.5 f 1 = 40 cm
Now, consider the combination of the two lenses with oil in between.
The system acts as a lens with the oil forming a biconvex lens.
The focal length of the oil lens is given by:
1 f 2 = ( n 2 − 1 ) ( 1 R − 1 − R ) 1 f 2 = ( 1.7 − 1 ) ( 1 20 + 1 20 ) 1 f 2 = 0.7 × 2 20 1 f 2 = 1.4 20 f 2 = 20 1.4 f 2 = 14.29 \\
\frac{1}{f_2} = (n_2 - 1) \left(\frac{1}{R} - \frac{1}{-R}\right) \\
\frac{1}{f_2} = (1.7 - 1) \left(\frac{1}{20} + \frac{1}{20}\right) \\
\frac{1}{f_2} = 0.7 \times \frac{2}{20} \\
\frac{1}{f_2} = \frac{1.4}{20} \\
f_2 = \frac{20}{1.4} \\
f_2 = 14.29 f 2 1 = ( n 2 − 1 ) ( R 1 − − R 1 ) f 2 1 = ( 1.7 − 1 ) ( 20 1 + 20 1 ) f 2 1 = 0.7 × 20 2 f 2 1 = 20 1.4 f 2 = 1.4 20 f 2 = 14.29 cm
The effective focal length
of the combination is given by:
1 F = 1 f 1 + 1 f 2 + 1 f 1 1 F = 1 40 + 1 14.29 + 1 40 1 F = 1 40 + 1 40 + 1 14.29 1 F = 2 40 + 1 14.29 1 F = 1 20 + 1 14.29 \\
\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} + \frac{1}{f_1} \\
\frac{1}{F} = \frac{1}{40} + \frac{1}{14.29} + \frac{1}{40} \\
\frac{1}{F} = \frac{1}{40} + \frac{1}{40} + \frac{1}{14.29} \\
\frac{1}{F} = \frac{2}{40} + \frac{1}{14.29} \\
\frac{1}{F} = \frac{1}{20} + \frac{1}{14.29} \\
F 1 = f 1 1 + f 2 1 + f 1 1 F 1 = 40 1 + 14.29 1 + 40 1 F 1 = 40 1 + 40 1 + 14.29 1 F 1 = 40 2 + 14.29 1 F 1 = 20 1 + 14.29 1 Calculate the sum:
1 F = 0.05 1 + 0.07 1 1 F = 0.12 F = 1 0.12 F = 8.33 \\
\frac{1}{F} = \frac{0.05}{1} + \frac{0.07}{1} \\
\frac{1}{F} = 0.12 \\
F = \frac{1}{0.12} \\
F = 8.33 F 1 = 1 0.05 + 1 0.07 F 1 = 0.12 F = 0.12 1 F = 8.33 cm
However, since the lenses are in contact and the oil lens is negative,
the effective focal length should be negative.
cm
Therefore, the focal length of the combination is
cm, which corresponds to Option 2.