PHY-111 LAB QUIZ PREPARATION

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Refractive Index of a Liquid using a Plane Mirror and a Convex Lens This experiment determines the refractive index ($\mu$) of a given liquid by forming a plano-concave liquid lens between a convex lens and a plane mirror. Key Concepts & Formulas 1. Refractive Index Definition For a certain pair of media and a ray of specific wavelength, the ratio of the sine of the angle of incidence ($i$) to the sine of the angle of refraction ($r'$) is a constant, known as the refractive index ($\mu$) of the second medium with respect to the first medium. $$ \mu = \frac{\sin i}{\sin r'} $$ 2. Lens Maker's Formula (for a lens in air) For a lens made of material with refractive index $\mu$ in air, with radii of curvature $r_1$ (first surface) and $r_2$ (second surface), its focal length $f$ is given by: $$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{r_1} - \frac{1}{r_2}\right) $$ For a bi-convex lens, $r_1$ is positive and $r_2$ is negative. For a plano-convex lens, one radius is infinite. 3. Plano-Concave Liquid Lens When a convex lens is placed on a drop of liquid on a plane mirror, the liquid forms a plano-concave lens. The curved surface of this liquid lens has the same radius of curvature ($r$) as the convex lens's bottom surface. The plane surface has an infinite radius of curvature. For a plano-concave lens, $r_1 = r$ (concave surface, so negative if light enters from liquid), and $r_2 = \infty$ (plane surface). Or, if we consider the liquid lens is effectively diverging, then $r_1 = r$ (convex surface of air-liquid interface) and $r_2 = \infty$. The formula for its focal length $f$ (for the liquid lens) becomes: $$ \frac{1}{f} = (\mu_{\text{liquid}} - 1)\left(\frac{1}{r} - \frac{1}{\infty}\right) = (\mu_{\text{liquid}} - 1)\frac{1}{r} $$ Rearranging for $\mu_{\text{liquid}}$: $$ \mu_{\text{liquid}} = 1 + \frac{r}{f} $$ Here, $r$ is the radius of curvature of the convex lens's surface that forms the liquid lens. 4. Combined Lens System When the convex lens (focal length $f_1$) is placed on the liquid layer (forming a plano-concave liquid lens with focal length $f$), they act as a combined lens with focal length $F$. $$ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f} $$ From this, the focal length of the liquid lens $f$ can be found: $$ \frac{1}{f} = \frac{1}{F} - \frac{1}{f_1} = \frac{f_1 - F}{Ff_1} \implies f = \frac{Ff_1}{f_1 - F} $$ Since the plano-concave liquid lens is a diverging lens, its focal length $f$ is negative. If we consider the absolute value of $f$ for calculation, then: $$ |f| = \frac{Ff_1}{|f_1 - F|} $$ Apparatus Spherometer (to measure radius of curvature $r$) Bi-convex lens Plane mirror Liquid (e.g., water, oil) Stand with clamp (to hold object/pin) Object (e.g., pen, pin) Meter scale Slide calipers (to measure lens thickness for $f_1$) Procedure Summary Part 1: Measure Radius of Curvature ($r$) of the Convex Lens Use a spherometer to measure the distance $a$ between the legs of the spherometer. Determine the pitch and least count of the spherometer. Place the spherometer on the convex lens and measure the sagitta $h$ (displacement of the middle leg) of the lens surface. Calculate $r$ using the formula: $r = \frac{a^2}{6h} + \frac{h}{2}$. Part 2: Measure Focal Length ($f_1$) of the Convex Lens Place the convex lens on a plane mirror. Adjust the object (pin) vertically above the lens until its real, inverted image (formed by reflection from the mirror after passing through the lens) coincides with the object itself. This occurs when the object is at the focal point. Measure the distance from the object to the lens using a meter scale. This is $f_1$. Alternatively, use the formula $f_1 = l_1 + t/3$, where $l_1$ is the distance from the pin to the top of the lens, and $t$ is the thickness of the lens. Part 3: Measure Focal Length ($F$) of the Combined Lens System Place a few drops of the liquid on the plane mirror. Carefully place the convex lens on the liquid drops, forming a plano-concave liquid lens. Repeat step 2 from Part 2 to find the focal length $F$ of this combined lens system. Alternatively, use the formula $F = l_2 + t/3$, where $l_2$ is the distance from the pin to the top of the combined lens, and $t$ is the thickness of the convex lens. Part 4: Calculate Refractive Index ($\mu$) of the Liquid Calculate the focal length $f$ of the plano-concave liquid lens using: $f = \frac{Ff_1}{f_1 - F}$. Remember $f$ will be negative for a diverging lens. Calculate the refractive index of the liquid using: $\mu = 1 + \frac{r}{|f|}$. Lab Quiz Questions *1) Define 'refractive index of a medium'. The refractive index of a medium is a dimensionless quantity that describes how fast light travels through the medium. It is defined as the ratio of the speed of light in a vacuum ($c$) to the speed of light in the medium ($v$), i.e., $\mu = c/v$. Alternatively, it is the ratio of the sine of the angle of incidence to the sine of the angle of refraction when light passes from one medium to another. *2) Draw a lens having radii of curvature $r_1$ and $r_2$. If $\mu$ is the refractive index of the material of the lens with respect to air, then write down the formula of its focal length, $f$. $r_1$ $r_2$ The focal length $f$ is given by the Lens Maker's Formula: $$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{r_1} - \frac{1}{r_2}\right) $$ *3) Draw a plano-concave lens. You can consider that the plane surface of this lens is a portion of a huge sphere, hence its radius of curvature is $\infty$. For a plano-concave lens by using the equation of question 2 prove that, $\mu = 1 + \frac{r}{f}$. Here $r$ is the radius of curvature of the concave surface of the lens. $r$ For a plano-concave lens, one surface is plane ($r_2 = \infty$) and the other is concave ($r_1 = r$). The formula should be adapted to the convention used. If we take $r_1 = r$ (concave surface) and $r_2 = \infty$ (plane surface), and for a diverging lens, $f$ is negative, then using the lens maker's formula: $$ \frac{1}{f} = (\mu - 1)\left(\frac{1}{-r} - \frac{1}{\infty}\right) = (\mu - 1)\left(-\frac{1}{r}\right) $$ $$ -\frac{1}{|f|} = -(\mu - 1)\frac{1}{r} $$ $$ \frac{1}{|f|} = (\mu - 1)\frac{1}{r} $$ $$ \mu - 1 = \frac{r}{|f|} $$ $$ \mu = 1 + \frac{r}{|f|} $$ Note: The sign convention for $r$ can be tricky. The derivation in the document implicitly assumes $r$ is positive and $f$ is treated as negative for a diverging lens, then its absolute value is used. 4) Please see the figure where a point object is kept on one of the principal foci, F1 and a mirror is kept on the other side of the lens, perpendicular to the principal axis. Where the image of the point object will form, explain with the help of ray diagrams. F1 F2 Explanation: An object is placed at the principal focus F1 of the convex lens. Ray 1 (blue): A ray originating from F1 and passing through the optical center (O) of the lens will go undeviated. It then strikes the plane mirror perpendicularly at F2. Ray 2 (red): A ray originating from F1 and traveling parallel to the principal axis after refraction by the convex lens will pass through the second principal focus (F2) of the lens. It then strikes the plane mirror at F2. Since both refracted rays strike the plane mirror perpendicularly (at F2), they will retrace their paths exactly after reflection. The rays will then pass through the lens again and converge back to the original object position F1. Therefore, the image of the point object will form at F1 (the original object position) . Spherometer Questions 1. What is the pitch of a spherometer? The pitch of a spherometer is the linear distance moved by the central screw for one complete rotation of the circular scale. 2. How do you find the least count (L.C.) of a spherometer? Least Count (L.C.) = $\frac{\text{Pitch}}{\text{Number of divisions on circular scale}}$. 3. Write down the formula for the radius of curvature ($r$) of a spherical surface using a spherometer readings $a$ and $h$. $r = \frac{a^2}{6h} + \frac{h}{2}$, where $a$ is the mean distance between any two outer legs, and $h$ is the sagitta (displacement of the central screw from the plane formed by the outer legs). Vernier Caliper Questions 1. What is the Vernier constant (least count) of a Vernier caliper? The Vernier constant (least count) is the difference between one main scale division (MSD) and one Vernier scale division (VSD). It is typically $0.01 \text{ cm}$ or $0.02 \text{ mm}$. Calculated as: $LC = 1 \text{MSD} - 1 \text{VSD} = \frac{\text{Value of 1 MSD}}{\text{Total number of divisions on Vernier scale}}$. 2. How do you find the total reading of a Vernier caliper? Total Reading = Main Scale Reading (MSR) + (Vernier Coincidence (VC) $\times$ Least Count (LC)) - Zero Error. 3. Explain positive and negative zero error in a Vernier caliper. Positive Zero Error: When the jaws are closed, if the zero mark of the Vernier scale is to the right of the zero mark of the main scale. It is subtracted from the total reading. Negative Zero Error: When the jaws are closed, if the zero mark of the Vernier scale is to the left of the zero mark of the main scale. It is added to the total reading.