Question

Two points separated by a distance of $0.1$ mm can just be inspected in a microscope when the light of a wavelength $600A^0$ is used. If the light of wavelength $4800A^0$ is used, the limit of resolution will become

• A. $0.80mm$
• B. $0.12mm$
• C. $0.10mm$
• D. $0.08mm$

Answer 1

Answer: (D)

$0.08mm$

Resolving power is given by:

$Resolvingpower\left(RP\right)=\frac{\lambda }{2a}$

Since the aperture of the microscope remains the constant, the resolving power can be given by:

$\therefore \frac{R{P}_1}{R{P}_2}=\frac{{\lambda }^{{}^{\prime }}}{\lambda }$

$R{P}_2=0.1\times \frac{4800}{6000}$

$R{P}_2=0.08mm$.
$\therefore$ Resolution improves when the light of the lower wavelength is used.