Optical Instruments MCQ Quiz - Objective Question with Answer for Optical Instruments - Download Free PDF

Concept:

The magnification of an astronomical telescope is given by the formula:

M = f_o / f_e,

where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece lens.

The tube length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece lens.

Explanation:

Given:

Focal length of the objective lens, f_o = 10 m = 1000 cm

Focal length of the eyepiece lens, f_e = 10 cm

Therefore, the tube length of the telescope is:

Tube length = f_o + f_e = 1000 cm + 10 cm = 1010 cm

The magnification of the telescope is:

Magnification (M) = f_o / f_e = 1000 cm / 10 cm = 100

The correct option is (4).

\(M\) = magnification of compound microscope

\(\Rightarrow M = M_0 \left(1 + \dfrac{D}{f_e}\right)\) ............(1)

where \(D\) = distance of nearest vision = \(25\) cm

\(f_e\) = focal length of eye lens

Given,

\(m = 100\)

\(m_0 = 10\)

\(D = 25\)

substituting in (1)

\(100 = 10 \left(1 + \dfrac{25}{f_e}\right)\)

\(f_e = \dfrac{25}{9}\) cm

Calculation:

The magnifying power (MP) of a simple microscope when the final image is formed at the least distance of distinct vision (D = 25 cm) is given by the formula:

MP = 1 + (D / f)

Where:

D = 25 cm (least distance of distinct vision),

f = 10 cm (focal length of the convex lens).

Substituting the given values:

MP = 1 + (25 / 10) = 1 + 2.5 = 3.5

Thus, the magnifying power of the simple microscope is 3.5.

Calculation:

In an astronomical telescope, the magnifying power (M) is given by the ratio of the focal length of the objective (f₀) to the focal length of the eyepiece (fₑ):

M = f₀ / fₑ

Given that the magnifying power is 7, we have:

7 = f₀ / fₑ

This implies:

f₀ = 7 fₑ

Also, it is given that the objective and eyepiece are 32 cm apart. The separation between the objective and the eyepiece in a telescope adjusted for normal vision is equal to the sum of their focal lengths:

f₀ + fₑ = 32 cm

Substituting the value of f₀ from the first equation f₀ = 7 fₑ into the second equation, we get:

7 fₑ + fₑ = 32

Combining like terms:

8 fₑ = 32

Solving for fₑ, we get:

fₑ = 32 / 8 = 4 cm

Now, substituting this back to find f₀:

f₀ = 7 × 4 = 28 cm

Therefore, the focal lengths of the objective and the eyepiece are:

f₀ = 28 cm

fₑ = 4 cm

Hence, the correct option is:

Option 4: f₀ = 28 cm and fₑ = 4 cm

Explanation:

The eye muscles, specifically the ciliary muscles, control the shape of the lens to focus on objects at different distances. When viewing distant objects, the ciliary muscles are relaxed, and the lens is in its least convex (thin) shape, reducing strain.

In contrast, when focusing on near objects, the ciliary muscles contract to make the lens more convex, increasing strain. The closer the object, the more strain is required.

Thus, when looking at faraway objects, the eye muscles experience the least strain.

Thus, option '1' is correct.

Compound microscope: 

• It is used for much larger magnifications.
• Two lenses are used with one lens compounding the effect of the other.

where fo is the focal length of the objective lens,  fe  is the focal length of the eyepiece, h is the object height, h' is the size of the first image

• A compound microscope uses two lenses i.e., objective lens and eyepiece. 
  • Objective lens: lens nearest to the object forming a real, inverted, and magnified image of the object.
  • Eyepiece: produces a final image that is enlarged and virtual.

​CALCULATION:

Given: magnification of compound microscope(m) = 20, focal length of eye piece (f_e) = 5 cm and image is formed at near point (v= D) = 25 cm.

• Linear magnification due to eyepiece:

\(⇒ m_{e} = 1+\frac{D}{f} = 1+ (\frac {25}{5} )= 6\)

• Total magnification is given by:

⇒ m = m_o × m_e,

where m_o is the magnification due to the objective lens.

• Magnification due to the objective lens is given by:

\(⇒ m_{o} = \frac {m}{m_{e}} = \frac {20}{6} = 3.33\)

• Hence option 3) is correct.

The correct answer is option 3) i.e. the product of the magnification of objective lens and eyepiece

CONCEPT:

• Compound microscope: A compound microscope is an optical instrument that is used to obtain highly magnified images. It uses a set of two lenses - the eyepiece lens and the objective lens.
  • Generally, the eyepiece used is of lower power than that of the objective lens.
  • A typical working of the compound microscope is as shown.

• Magnification of a compound microscope (m) is given by the formula

m = mo × me

Where mo is the magnifying power of the objective lens and me is the magnifying power of the eyepiece.

EXPLANATION:

Magnification of a compound microscope, m = mo × me

• Therefore, the magnification of a compound microscope is the product of the magnification of the objective lens and eyepiece.

The correct answer is Convex lens.

Key Points

CONCEPT:

• Magnifying glass: It is an optical instrument used to obtain a magnified image of an object. It consists of a simple convex lens that is housed in a frame along with a handle.
  • When light reflects from an object, it travels parallel to each other.
  • When they pass through a magnifying glass, the convex lens converges the parallel rays and forms an enlarged virtual image on the retina.
  • The converged rays if extended give the impression of a large image formed, this is how magnification occurs.

EXPLANATION:

• A magnifying lens is used to obtain magnified images of an object placed in front of us. This is possible only if we can look at the object through transparent material. Hence, it uses a lens.
• A concave lens is diverging in nature and will not cause image formation on the retina. A convex lens is converging in nature and therefore causes the light rays from the object to converge at the retina. 
• Therefore, a magnifying glass comprises a simple convex lens.

Concept:

Resolving power is defined as the inverse of the distance or angular separation between two objects which can be just resolved when viewed through the optical instruments.

According to Rayleigh criterion

In telescope

angular separation b/w two object

\({\rm{\Delta }}\theta = 1.22\frac{\lambda }{d}\)

Resolving power \( = \frac{1}{{{\rm{\Delta }}\theta }} = \frac{d}{{1.22\lambda }}\)

d = diameter

λ = wavelength

In microscope

Distance between two object

\({\rm{\Delta }}d = \frac{\lambda }{{2n\sin \theta }}\)

Resolving power \( = \frac{1}{{{\rm{\Delta }}d}} = \frac{{2n\sin \theta }}{\lambda }\)

n = refractive index of medium separating object and aperture

Calculations:

Given,

Wavelength = l = 250 nm = 2.5 × 10^-7 m

Aperture = D = 0.5 m

We know,

The resolving power of the telescope =

\(\frac{D}{1.22\times \lambda}=\frac{0.5}{1.22\times 2.5 \times 10^{-7}}=1.64\times10^{6}\)

The correct answer is Helioscope.

Key Points

• Helioscope was first used by Benedetto Castelli in 1578 
• Since observing the sun directly can damage eyesight and its difficult to observe anything directly in Helioscope sunlight is projected on a white sheet of paper with a help of telescope-like equipment and then any changes like eclipse can be observed easily by observing the sheet 

Explanation:

• A helioscope is an instrument used for observing the sun and sunspots.
• The helioscope was first used by Benedetto Castelli and refined by Galileo Galilei.

CONCEPT:

• Power of Lens: The inverse of the focal length is known as the power of the lens.
  • It shows the bending strength for the light ray of the lens.
  • The unit of power of a lens is Dioptre when the focal length of the lens is taken in meter (m).

\(P = \frac{1}{f\,(m)}=\frac{100}{f\,(cm)}\)

where P is the power of the lens and f is the focal length of the lens.

EXPLANATION:

• From the above, it is clear that the unit of power of a lens is Dioptre. Therefore option 3 is correct.

NOTE:

• Concave lens: It is a diverging lens that diverges the parallel beam of light.
  • It can also gather light from all directions and project it as a parallel beam.
  • The focal length of the concave lens is negative.
  • It has a virtual focus from the diverging rays of light that seem to converge.
• Convex lens: The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.
  • The focal length of a convex lens is positive.

CONCEPT:

Microscope:

• A microscope is an optical instrument that forms a large image of close and tiny objects.

Resolving power:

• The resolving power of a microscope is the inverse of the shortest distance between two separate points in a microscope's field of view that can be seen distinctly.

Resolving power is given as,

\(⇒ Resolving\,power=\frac{1}{Δ d}= \dfrac{2a}{λ}\)

Where Δd = shortest distance between two separate points, a = aperture, and λ = wavelength

CALCULATION:

• Resolving power of the microscope is given as,

\(⇒ Resolving\,power=\frac{1}{Δ d}= \dfrac{2a}{λ}\)

\(⇒ Resolving\,power\propto \dfrac{1}{λ}\)     -----(1)

• We know that the relation of the wavelength of the red light, violet light, infrared light, and the visible light is given as,

⇒ λ_Infrared > λred > λvisible > λviolet

• By equation 1 it is clear that the resolving power of the microscope is inversely proportional to the wavelength of the light, so the resolving power for the violet light will be maximum because the violet light has a minimum wavelength.
• Hence, option 2 is correct.

Additional Information

• ​Infrared Rays: Infrared radiation (IR), is an electromagnetic wave with wavelengths longer than visible light.
  • Infrared rays are undetectable by the human eye, although IR of wavelengths up to 1050 nanometers from specially pulsed lasers can be seen by humans under certain conditions.
  • Infrared light extends from the suggested red edge of the visible spectrum at 700 nanometers to 1 millimeter.
  • Infrared is commonly divided into five categories as near-wavelength, short-wavelength, mid-wavelength, long-wavelength, and far-infrared.
• ​Visible spectrum: The portion of the electromagnetic spectrum that is visible to the human eye is known as the visible light spectrum.
  • The visible light range comes under the range of electromagnetic spectrum between infrared and ultraviolet having a frequency of about 4 x 10 14 to 8 x 10 14 cycles per second and wavelengths of about 740 nm or 2.9 x10 -5 inches to 380 nm.

Concept:

• Optical fibres are transparent fibres and act as a light pipe to transmit light between its two ends. They are made up of silicon dioxide.

• The total internal reflection occurs when the angle of incidence is greater than the critical angle.
• In optical fibre glasses of high and lower index are assembled in the precise order.
• The light comes in from one end of the fibre and after thousands of successive internal reflections, the light reaches the opposite end of the fibre with almost zero loss. 

Explanation:

• Since in case of optical fibre, there is a total internal reflection of light. So the basic principle of optical fibre is Total internal reflection.

CONCEPT:

• Power of Lens: The inverse of the focal length is known as the power of the lens.
  • It shows the bending strength for the light ray of the lens.
  • The unit of power of a lens is Dioptre when the focal length of the lens is taken in meter (m).

\(P = \frac{1}{f\,(m)}=\frac{100}{f\,(cm)}\)

where P is the power of the lens and f is the focal length of the lens.

• Concave lens: It is a diverging lens that diverges the parallel beam of light.
  • It can also gather light from all directions and project it as a parallel beam.
  • The focal length of the concave lens is negative.
  • It has a virtual focus from the diverging rays of light that seem to converge.
• Convex lens: The lens whose refracting surface is upside is called a convex lens.
  • The convex lens is also called a converging lens.
  • The focal length of a convex lens is positive.

CALCULATION:

Given - Power of lens (P) = + 2.5 D

• The power of a lens is given by

\(\Rightarrow f = \frac{{100}}{{{P}}} = \frac{{100}}{{ 2.5}} = 40\,cm\)

• As the focal length is positive, therefore the lens is convex and has a focal length of 40 cm.

Concept:

In an optical microscope, a lens is used to see objects.

The resolving power of a lens is the power of a lens to differentiate between objects.

The larger the power of the lens smaller will be the minimum distance between the lines that can differ from one another.

Resolving power depends upon the wavelength.

Resolving power ∝ \(\frac{1}{\lambda}\)

where λ is the wavelength of radiation.

Calculation:

Wavelengths are λ₁ = 4000 Å and λ₂ = 6000 Å is

Resolving power ∝ \(\frac{1}{\lambda}\)

\(\frac{R_1}{R_2} = \frac{\lambda_2}{\lambda_1}\)

\(=\frac{6000Å}{4000Å}\)

\(= \frac{3}{2}\)