MCQ
Resolving power of a microscope depends upon
- AWavelength of light used, directly
- ✓Wavelength of light used, inversely
- CFrequency of light used
- DFocal length of objective
✓
Answer
Correct option: B.
Wavelength of light used, inversely
b
( b)Resolving power of microscope $ \propto \frac{1}{\lambda }$
( b)Resolving power of microscope $ \propto \frac{1}{\lambda }$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
An electron of charge ‘e’ coulomb passes through a potential difference of V volts. Its energy in ‘joules’ will be
In an electron gun, the electrons are accelerated by the potential $V$. If e is the charge and $m$ is the mass of an electron, then the maximum velocity of these electrons will be
→In the given circuit calculate potential difference between $A$ and $B$ ............... $\mathrm{V}$
→A solenoid has $2000$ turns wound over a length of $0.3\,m$. The area of cross-section is $1.2\times10^{-3}\,m^2$. Around its central section a coil of $300$ turns is closely wound. If an initial current of $1\,A$ is reversed in $0.25\,s$. Find the emf induced in the coil.......$mV$
→1 mg gold undergoes decay with 2.7 days half-life period, amount left after 8.1 days is
The charge distribution along the semi-circular arc is non-uniform . Charge per unit length $\lambda $ is given as $\lambda = {\lambda _0}\sin \theta $ , with $\theta $ measured as shown in figure. $\lambda_0$ is a positive constant. The radius of arc is $R$ . The electric field at the center $P$ of semi-circular arc is $E_1$ . The value of $\frac{{{\lambda _0}}}{{{ \in _0}{E_1}R}}$ is
→A sinusoidal ac current flows through a resistor of resistance $R$. If the peak current is${I_p}$, then the power dissipated is
→A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay $X_Z^A \rightarrow Y_{Z+1}^A+\beta^{-}+\bar{v}_e$ will occur is ( $m_e$ denotes the mass of the $\beta$ particle and the neutrino mass $m_v$ can be neglected)
→A parallel plate capacitor having capacitance $12\, pF$ is charged by a battery to a potential difference of $10\, V$ between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant $6.5$ is slipped between the plates. The work done by the capacitor on the slab is.......$pJ$
→Complete the following statement. For electromagnetic induction to occur: