(A) an oil with refractive index of 1.6
(B) an oil with refractive index of 1.5
(C) an oil with refractive index of 1.4
(D) an oil with refractive index of 1.3
Oil immersion objectives achieve maximum resolution when the oil’s refractive index matches glass closely. This article solves Q.75 for 100× objectives with the correct answer and all options explained.
Correct Answer
The correct answer is (A) an oil with refractive index of 1.6.
Resolution follows d = λ/(2NA), where NA = n(sinα); higher oil refractive index n captures more oblique rays, maximizing NA up to 1.4-1.6. Oils near glass RI (1.51-1.52) minimize refraction loss; 1.6 provides superior light gathering vs lower values.
Option Breakdowns
(A) an oil with refractive index of 1.6
Highest RI captures widest light cone through specimen, yielding maximum NA and sub-micron resolution.
Ideal for advanced apochromatic 100× objectives.
(B) an oil with refractive index of 1.5
Standard commercial oil (n≈1.515) matches glass coverslips well but transmits fewer oblique rays than 1.6.
Good resolution but suboptimal vs higher RI.
(C) an oil with refractive index of 1.4
Lower RI increases refraction at oil-glass interface, reducing effective NA and resolution.
Compromises image clarity for 100× oil objectives.
(D) an oil with refractive index of 1.3
Approaches air (n=1.0) performance; significant light loss and spherical aberration.
Unsuitable for high-NA oil immersion microscopy.
| Option | RI Value | NA Impact | Resolution Quality |
|---|---|---|---|
| (A) 1.6 | Highest | Maximum light capture | Best |
| (B) 1.5 | Standard | Good matching | Adequate |
| (C) 1.4 | Moderate | Reduced oblique rays | Fair |
| (D) 1.3 | Low | Major refraction loss | Poor |
