Sessional Lab - Optics

Experiment E10 (L1): Refractive Index of a Liquid by using Lens and a Plane Mirror

Q1: What is the nature of the liquid lens? Ans: It is a plano-concave lens.

Q2: Is the focal length of the liquid lens greater or smaller than that of the convex lens? Ans: Greater, otherwise the combination cannot behave as a convex lens having the first principal focus on the positive side.

Q3: Can the refractive index of any value be measured by this method? Ans: No, when the refractive index of the liquid is much less than the focal lengths of both lenses are equal, the method fails. The maximum refractive index of the liquid is $μ = 1 + \frac{r}{f _{2}}$ , and then $μ = 1 + 2$ (for $f_{2} = r$ ).

Q4: Can you find the refractive index of the given liquid without using a spherometer? Ans: Yes, by repeating the experiment with a liquid of known refractive index ( $μ^{'}$ ), the focal length $f_{2}^{'}$ of the liquid concave lens is found out. Then, $μ = 1 + \frac{r}{f _{2}}$ is derived from $r = (μ^{'} - 1) f_{2}^{'}$ . Hence the refractive index of the given liquid is $μ = 1 + \frac{( μ ^{'} - 1 ) f _{2}^{'}}{f _{2}}$ .

Q5: What is parallax error and how to avoid it? Ans: Parallax error exists when two objects are not coincident, which is detected by their relative displacement between them when the eye is moved. When there is no relative displacement, there will be no parallax.

Q6: Can you perform this experiment by a concave lens? Ans: No, because a concave lens cannot produce any real image.

Experiment E11 (L2): Focal Length by Displacement Method

Q1: Why do you keep the distance between the object and the screen greater than 4f? Ans: For getting two images, the distance between the object and the screen should be greater than $4 f$ . Otherwise, two images cannot be obtained.

Q2: What is the minimum distance between the object and its real image formed by a convex lens? Ans: The minimum distance between the object and its real image is equal to $4 f$ . If the distance between the object and the image is less than $4 f$ , two images cannot be formed. Then, the displacement method cannot be used for the determination of the focal length of the lens.

Q3: Is it advisable to make $d$ large? Ans: No, because the diminished image will be very small, and its focused condition cannot be traced.

Q4: Why the index correction for $x$ is not necessary? Ans: For the displacement of the lens, it must be equal to the displacement of the index mark of the lens stand.

Q5: What is the advantage of this method? Ans: No index error for $x$ is necessary. Thickness of the lens is not required.

Q6: Can the size of the object be obtained by measuring the sizes of the two images? Ans: Yes, if $I_{1}$ and $I_{2}$ are the sizes of the two images, then the size of the object is expressed as $O = I_{1} I_{2}$ .

Q7: How to test a lens whether it is a concave convex or mirror? Ans: If the lens is placed very close to a painted paper, then if the image is: i) Magnified, the lens is convex. ii) Diminished, the lens is concave. iii) Equal to that of the object, it is a mirror.

Q8: Why one image is magnified while the other diminished? Ans: Magnification is defined as $\frac{image distance}{object distance}$ . When the lens is nearer to the object, the image distance is greater and the image is magnified. When the lens is nearer to the screen, the image distance is small and the object distance is greater, and the inverted image is diminished.

Q9: By employing your data can you find ‘f’ graphically? Ans: Yes. The final equation can be written as $x^{2} / d = d - 4 f$ . If a graph is plotted with $x^{2} / d$ as the ordinate and $d$ as the abscissa, a straight line will be obtained and its intercept on the x-axis, i.e., $x^{2} / d = 0$ , will give the value of $4 f$ numerically.

Q10: Under what condition a real magnified or a diminished image be formed? Ans: When the focus ( $f$ ) and $2 f$ are the distance between the object and the lens:

When the object is placed between $f$ and $2 f$ , the image will be real and magnified.
But if the object is placed between $2 f$ and infinity, the image will be real and diminished.

Q11: What is index rod? Ans: An index rod is a rod mounted on a stand with pointed ends.

Q12: Why index correction is necessary? Ans: In optical benches, the readings are usually taken on the scale at the base end. The distance between two objects on the principal axes may not be the same as the distance of the index at their base on the bench scale.

That is an important apparatus! According to your source material for Experiment E12, the Spectrometer is used to measure the refractive index by finding the angle of the prism and the angle of minimum deviation.

Let’s continue gathering the questions and answers for the next five experiments, focusing on the Light section, which includes several spectrometer experiments: E12, E13, E14, E15, and E16.

Experiment E12 (L3): Refractive Index of the Material of a Prism using a Spectrometer

Q1: Why the spectrometer should be leveled before its use? Ans: Otherwise the position of the image will change with the position of the telescope, i.e., it will be different for different positions of the telescope.

Q2: What the Spectrometer and how are the conditions for obtaining a pure spectrum realized in it? Ans: The Spectrometer is used to measure the refractive index by finding the angle of the prism and the angle of minimum deviation. If the telescope of the spectrometer is replaced by a photographic camera, the instrument is transformed into a spectrograph. The measurement of refractive indices of a material for various spectral lines is provided by the spectrometer, which is a compact apparatus for obtaining a pure spectrum. It is equipped with a circular scale and a Vernier.

Q3: What are the main parts of a spectrometer? Ans: In its simplest form, it has the following main parts: i) the collimator, (ii) the telescope, and (iii) the prism table.

Q4: What is the function of the collimator? Ans: Collimator produces parallel beam of light.

Q5: Why do you set the telescope for parallel rays? Ans: If the incident rays are not parallel, i.e., if they are either divergent or convergent, the image focused for one position of the prism will be out of focus for other positions. If the incident rays are parallel, then both the object and the image of the prism will be at infinity, and the telescope, once focused for the image, will remain so for every position of the prism.

Q6: How do you place the prism when determining its angle by rotating the telescope? Ans: The prism table should be placed with its edge over the center of the prism table.

Q7: How do you place the prism when determining its angle by rotating the prism table? Ans: The prism table should be placed with its refracting edge at the center of the prism table when reading the angle by rotating the telescope.

Q8: How does the deviation change with color of light? Ans: Deviation is greater for violet than for red light. If the difference of wave length of light is considered, deviation will be different.

Q9: Why do you take sodium light and not white light? Ans: Sodium light is taken because it gives a single image of the slit and a single value of minimum deviation, while white light gives a spectrum and the value of minimum deviation is different for different wavelengths of light.

Q10: What is the condition of minimum deviation? Ans: At the minimum deviation position, the angle of incidence should be equal to the angle of emergence.

Q11: How does the deviation change with the color of light? Ans: Deviation increases with shorter wavelengths, e.g., it is greater for violet light than for red light.

Q12: How does the deviation of a ray vary with its angle of incidence? Ans: The deviation has becomes minimum at a particular angle of incidence, but it always increases when the angle of incidence is either increasing or decreasing than that at which the deviation becomes minimum.

Q13: What do you mean by monochromatic light? Is sodium light strictly monochromatic? Ans: Sodium light is strictly monochromatic. Light of a particular wavelength is called monochromatic. Sodium light is not exactly monochromatic, but it is considered a monochromatic light as it contains light of two wavelengths of values $589 0^{\circ} A$ and $589 6^{\circ} A$ .

Q14: To measure the angle of a thick prism why the edge of the prism is placed at the center of the table? Ans: The error in the measurement of the angle of a thick prism is minimal by such placing.

Q15: Why readings of two Verniers are noted? Ans: Both Verniers should be read to avoid error due to the non-coincidence of the center of the circular scale with the axis of rotation of the telescope or the table.

Q16: What kind of an image does the telescope produce? Ans: A virtual image at infinity is produced by the telescope. A real diminished image is produced by the objective of the telescope while the eyepiece produces a virtual magnified image.

Experiment E13 (L4): Wavelength of Sodium Light by measuring the Diameter of Newton’s Rings

Q1: What is the purpose of the traveling microscope? Ans: The original source states: See the text. (Self-Generated Answer based on theory): The traveling microscope is used for finding the length of very small objects. In this experiment, it is used to precisely measure the diameter of the Newton’s Rings.

Q2: What is the cause of Newton’s rings? Ans: The original source states: See the text. (Self-Generated Answer based on theory): Newton’s rings are caused by the interference between the light rays reflected from the lower surface of the lens and the upper surface of the optically plane glass plate. This interference results from the air film of varying thickness between the two surfaces.

Q3: Why are the rings circular? Ans: The original source states: See the text. (Self-Generated Answer based on theory): The rings are circular because the air film enclosed between the plane glass plate and the convex lens has a thickness that is constant for all points lying on a circle around the central point of contact.

Q4: Why does the center of the ring pattern appear dark in reflected light? Ans: The original source states: See the text. (Self-Generated Answer based on theory): The center is dark because the light reflecting off the lower surface of the lens (denser medium) undergoes a phase change of $π$ (or $18 0^{\circ}$ ) relative to the light reflected off the upper surface of the glass plate (rarer medium). At the point of contact, the air film thickness is zero ( $t = 0$ ), resulting in destructive interference and thus a dark spot.

Q5: Why is monochromatic light used? Ans: The original source states: See the text. (Self-Generated Answer based on theory): Monochromatic light (like Sodium light) is used because interference fringes (rings) formed by white light would overlap after the first few rings, making accurate measurement impossible due to the different wavelengths present.

Experiment E14 (L5): Wavelength of Sodium Light and the Resolving Power of a Plane Diffraction Grating using Spectrometer

Q1: What is Diffraction of light? Ans: Diffraction is the bending of light from the edge of a slit and spreading of light waves or the illumination of the geometrical shadow of an object.

Q2: What is the difference between interference and diffraction? Ans: In interference, the resultant intensity at a point is the resultant of the superposition of two wave fronts coming from two coherent sources. But in diffraction phenomenon, the resultant intensity at a point is due to the superposition of wavelets from two parts of a single wavefront.

Q3: What is diffraction grating? Ans: A diffraction grating is a device which consists of a very large number of closely spaced parallel slits separated by equal opaque spacing.

Q4: What is transmission and reflection grating? Ans:

Transmission grating: A plane diffraction grating consists of a very large number of narrow slits side by side. The slits are separated by opaque spaces. When a wave front is incident on a grating surface, light is transmitted through the slits and obstructed by the opaque portions. Such a grating is called the transmission grating.
Reflection grating: If the lines are drawn on a silvered surface (plane or concave), then the light is reflected from the positions of the mirror in between any two lines, and such surfaces act as reflection gratings.

Q5: What is grating element, grating constant and corresponding points? Ans:

Grating element: If $a$ be the width of a slit and $b$ be the width of an opaque space of a grating, then $(a + b)$ is called the grating element.
Grating constant: The reciprocal of the grating element, $(1/ (a + b))$ , is known as the grating constant.
Corresponding points: When two points in the consecutive free separated by a distance $(a + b)$ , the grating element, and then these two points are known as corresponding points.

Q6: What happens if the number of rulings per cm, $N$ , is increased or decreased? Ans: If $N$ is increased, the order number will be few but they will be separated by a large angle. If $N$ is decreased, the order number will be large, but separated by a small angle.

Q7: What is an absent spectrum? Ans: In diffraction grating the condition for principle maximum is $(a + b) sin θ = nλ$ . Hence the first order spectrum is absent if $λ > a + b sin θ$ and $sin θ > 1$ . The second order spectrum is absent if $(a + b) < 2 λ$ . In general if $(a + b) < nλ$ , then the $n^{t h}$ order spectrum will be absent.

Q8: What is the condition that the second order spectrum may be absent? Ans: In order that the $2^{n d}$ order spectrum may be absent, $\frac{a + b}{a} = 2$ , i.e., $a = b$ . In this case the width of the slit is equal to the opaque spacing between any two consecutive slits. In this case the $2^{n d}$ order spectrum will be absent.

Q9: What happens if the ruled surface of the grating is directed towards the collimator? Ans: The incident rays will be first diffracted at the ruled surface and then will be again refracted by the glass surface. Thus the angle $θ$ is not due to diffraction alone; it will be due to diffraction and refraction.

Q10: What do you mean by ghost line? Ans: If the rulings on a grating are not exactly equidistant or accurately parallel, then some additional lines appear near the main spectral lines. The additional lines are called ghost lines.

Q11: How does a grating spectrum differ from a prismatic spectrum? Ans: The angle of diffraction in the grating spectrum is proportional to $λ$ when $θ$ is not very large. But in the case of the prismatic spectrum, the violet end is more drawn out than the red end. Hence the grating spectrum is a pure spectrum than a prismatic spectrum.

Q12: What is replica grating? Ans: Replica gratings are constructed from the original grating, possibly prepared by contact printing on a fine grained photo graphic plate. The original grating surface a thin layer of collodion solution is poured and the solution is allowed to harden. Then, the film of the collodion is removed from the grating surface and then fixed between two glass plates. This serves as a plane transmission grating. This is called replica grating.

Q13: What do you mean by resolving power? Ans: The ability of an optical instrument, expressed in numerical measure, to resolve the images of two nearby points is termed as its resolving power. Resolving power is the ability of the instrument to show nearby spectral lines as separate ones.

Experiment E15 (L6): Wavelength of various Spectral Lines of Discharge Tube by using Spectrometer and a Plane Diffraction Grating

Q1: What is the grating element and grating constant? Ans: The original source states: Same as experiment number 5. (Self-Generated Answer based on E14/L5): The grating element is the sum of the width of a slit ( $a$ ) and the width of an opaque space ( $b$ ), or $(a + b)$ . The grating constant is the reciprocal of the grating element, $1/ (a + b)$ .

Q2: How does the angle of deviation change with the wavelength of light? Ans: The original source states: Same as experiment number 5. (Self-Generated Answer based on E14/L5): In a grating spectrum, the angle of diffraction ( $θ$ ) is proportional to the wavelength ( $λ$ ) when $θ$ is not very large, according to the grating equation $nλ = (a + b) sin θ$ .

Q3: What determines the order of the spectrum? Ans: The original source states: Same as experiment number 5. (Self-Generated Answer based on E14/L5): The order of the spectrum, denoted by $n$ , corresponds to the integer value in the grating equation $nλ = (a + b) sin θ$ . $n = 1$ is the first order, $n = 2$ is the second order, and so on.

Experiment E16 (L7): Specific Rotation of Sugar Solution by using a Polarimeter

Q1: What is polarization of light? Ans: Polarization of light is the process by which light vibrations are confined to one particular direction.

Q2: What is plane polarized light? Ans: When light is passed through crystals like tourmaline, quartz, etc., it acquires the property of one-sidedness. This means the vibration takes place only in a plane. Such a light is said to be plane polarized.

Q3: What is plane of polarization and plane of vibration? Ans:

Plane of polarization: The plane in which no vibrations occur.
Plane of vibration: The plane in which a vibration occurs. This plane is at right angles to the plane of polarization.

Q4: What is a polarimeter? What are the main parts of polarimeter? Ans: A polarimeter is an instrument used for measuring the angle of rotation of the plane of polarization by an optically active substance. The main parts of the polarimeter are two Nicol prisms working as polarizer and analyzer, and a glass tube between them.

Q5: What do you mean by optical activity, optical rotation and angle of rotation? Ans:

Optical activity: The property of rotating the plane of vibration of plane polarized light about its direction of travel by some crystal is known as optical activity.
Optical rotation: This phenomenon is known as optical rotation.
Angle of rotation: The angle through which the plane of polarization is rotated is known as the angle of rotation.

Q6: Define specific rotation. Ans: Specific rotation for the given wavelength of light at a given temperature is defined as the rotation produced by one decimeter ( $10 cm$ ) length of the solution of unit concentration.

Q7: On what factor does specific rotation depend? Ans: Specific rotation depends upon: i) Thickness of the medium. ii) Concentration of the solution or the density of the active substance in the solvent. iii) Wavelength of light. iv) Temperature.

Q8: Does angle of rotation and specific rotation depend on strength of sugar solution and length of the tube? Ans: Angle of rotation is proportional to length and concentration, but specific rotation is independent of these factors.

That’s the final push! Let’s gather the questions for your last four experiments (E17, E18, E19, and E20) to complete your preparation guide.

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