🔑:## Step 1: Determine the lensmaker's equation for a plano-convex lens in air.The lensmaker's equation is given by ( frac{1}{f} = (n-1) left( frac{1}{R_1} - frac{1}{R_2} right) ), where ( n ) is the refractive index of the lens material, and ( R_1 ) and ( R_2 ) are the radii of curvature of the two surfaces. For a plano-convex lens, ( R_2 = infty ), simplifying the equation. However, since the problem involves the lens being in contact with water, we need to consider the effective focal length when the lens is immersed in water.## Step 2: Calculate the effective focal length of the lens when placed next to the water-filled aquarium from the outside.When the lens is placed next to the aquarium from the outside, the object is in water, and the image is in air. We need to consider the refraction at the water-air interface and the lens itself. The effective focal length ( f_{eff} ) can be calculated using the formula ( frac{1}{f_{eff}} = frac{n_{water}}{f} ) because the lens's focal length is given in air, and we are considering its effect in water.## Step 3: Apply the lens equation to find the image position when the lens is placed from the outside.The lens equation is ( frac{1}{f} = frac{1}{s'} - frac{1}{s} ), where ( s ) is the object distance, and ( s' ) is the image distance. We need to adjust this equation for the case where the object is in water and the image is in air, considering the refractive indices.## Step 4: Calculate the image distance and enlargement when the object distance is 2.5f for the lens placed from the outside.Given ( s = 2.5f ), we can substitute into the adjusted lens equation to find ( s' ). The magnification ( M ) is given by ( M = -frac{s'}{s} ).## Step 5: Repeat the calculation for the case where the lens is placed from the inside of the aquarium.When the lens is placed from the inside, the object is in water, and the image is also in water. The effective focal length and the lens equation need to be adjusted accordingly to account for the water on both sides of the lens.## Step 6: Calculate the effective focal length of the lens when placed inside the aquarium.The effective focal length in water ( f_{water} ) is given by ( f_{water} = frac{f}{frac{n_{lens}}{n_{water}} - 1} ) because the lens is now immersed in water.## Step 7: Apply the lens equation to find the image position when the lens is placed from the inside.Using the effective focal length in water, we apply the lens equation ( frac{1}{f_{water}} = frac{1}{s'} - frac{1}{s} ) to find the image distance when the object distance is 2.5f.## Step 8: Calculate the image distance and enlargement for the object distance of 2.5f when the lens is placed from the inside.Substitute ( s = 2.5f ) into the adjusted lens equation to find ( s' ), and then calculate the magnification.## Step 9: Execute the calculations for the outside placement.First, calculate the effective focal length when the lens is in air, which is simply ( f ) since the given ( f ) is for the lens in air. Then, apply the lens equation considering the refraction indices for the water-air interface.## Step 10: Execute the calculations for the inside placement.Calculate the effective focal length of the lens in water using ( f_{water} = frac{f}{frac{1.50}{1.33} - 1} ), then apply the lens equation.## Step 11: Calculate ( f_{water} ) for the inside placement.( f_{water} = frac{f}{frac{1.50}{1.33} - 1} = frac{f}{1.125 - 1} = frac{f}{0.125} = 8f ).## Step 12: Apply the lens equation for the inside placement with ( s = 2.5f ).( frac{1}{8f} = frac{1}{s'} - frac{1}{2.5f} ).## Step 13: Solve for ( s' ) in the inside placement scenario.( frac{1}{8f} = frac{1}{s'} - frac{1}{2.5f} ) simplifies to ( frac{1}{s'} = frac{1}{8f} + frac{1}{2.5f} ).## Step 14: Calculate the common denominator and solve for ( s' ).( frac{1}{s'} = frac{2.5 + 8}{20f} = frac{10.5}{20f} ), thus ( s' = frac{20f}{10.5} ).## Step 15: Calculate the magnification for the inside placement.( M = -frac{s'}{s} = -frac{20f}{10.5} div 2.5f ).The final answer is: boxed{1.905}

🔑:To determine the maximum energy a single photon can have and its relation to the Planck length and quantum gravity, we must delve into the realms of quantum mechanics and general relativity, particularly focusing on the Planck scale. The Planck scale is a set of units of measurement that are based on the fundamental physical constants and represent the scale at which quantum gravity effects become significant.## Step 1: Understanding the Planck UnitsThe Planck units are defined using the gravitational constant (G), the speed of light (c), and the reduced Planck constant (ħ). The Planck length (lp), time (tp), and mass (mp) are given by:- (l_p = sqrt{frac{hbar G}{c^3}})- (t_p = sqrt{frac{hbar G}{c^5}})- (m_p = sqrt{frac{hbar c}{G}})These units represent the scale at which quantum gravity effects are expected to become significant.## Step 2: Calculating the Planck EnergyThe Planck energy (Ep) is the energy associated with the Planck mass (mp) and can be calculated using the formula:- (E_p = m_p c^2 = sqrt{frac{hbar c^5}{G}})This energy represents the scale at which quantum gravity effects are expected to become important.## Step 3: Relating Photon Energy to the Planck EnergyThe energy of a photon (E) is given by:- (E = hbar omega = frac{hc}{lambda})where (omega) is the angular frequency, (lambda) is the wavelength, h is the Planck constant, and c is the speed of light.## Step 4: Maximum Photon Energy ConsiderationFor a photon to have the maximum energy related to the Planck scale, its energy should approach the Planck energy. However, the concept of a "maximum" energy for a photon in the context of quantum gravity is more nuanced. The Planck energy is a theoretical limit beyond which our current understanding of physics, particularly quantum mechanics and general relativity, is expected to break down.## Step 5: Implications and DiscussionThe Planck energy ((E_p = sqrt{frac{hbar c^5}{G}})) is approximately (1.22 times 10^{18}) GeV. This energy scale is far beyond what current particle accelerators can achieve, and it represents a limit where quantum gravity effects are expected to become significant. A photon with energy approaching the Planck energy would have a wavelength approaching the Planck length ((l_p approx 1.62 times 10^{-35}) meters), which is the scale at which the smoothness of spacetime is expected to give way to quantum fluctuations.## Step 6: Conclusion on Maximum Photon EnergyGiven the theoretical nature of the Planck scale and the limitations of our current understanding of quantum gravity, it's challenging to assign a strict "maximum" energy to a photon without considering the context of the theory (e.g., quantum field theory in curved spacetime, loop quantum gravity, string theory). However, the Planck energy provides a theoretical benchmark for the scale at which quantum gravity effects become important.The final answer is: boxed{1.22 times 10^{18}}

🔑:## Step 1: Define the problem and the random variable TWe are given a random variable X and i.i.d. random variables X1, X2, ... . The random variable T represents the number of tries needed until the experiment is terminated, which happens when X is surpassed or repeated.## Step 2: Determine the probability distribution of TFor a finite set, let's denote the possible values of X as {x1, x2, ..., xn} with probabilities p1, p2, ..., pn. The probability that Xi takes the value xj is pj. T is a geometric random variable with parameter p = P(Xi ≥ X), since the experiment ends when Xi ≥ X for the first time.## Step 3: Calculate the expectancy of T for the finite set caseThe expectancy of T, E[T], can be calculated using the formula for the expectancy of a geometric distribution: E[T] = 1 / p, where p is the probability of success (in this case, surpassing or repeating X).## Step 4: Consider the continuous but not absolutely continuous caseIf X is continuous but not absolutely continuous, it has a mixed distribution with both discrete and continuous parts. The discrete part can be treated similarly to the finite set case, but the continuous part introduces complexity. The expectancy of T would depend on the specific distribution of X and how the continuous part affects the probability of surpassing or repeating X.## Step 5: Analyze the effect of the continuous part on E[T]For the continuous part, the probability of exactly repeating X is 0, so the experiment would terminate when Xi surpasses X. The expectancy of T would be influenced by the rate at which the cumulative distribution function (CDF) of X increases, as this affects the probability of surpassing X at each step.## Step 6: Conclude on the calculation of E[T] for the continuous caseGiven the complexity introduced by the continuous part, especially if X is not absolutely continuous, a general formula for E[T] might not be straightforward without more specific information about the distribution of X. However, for the discrete part or when X is finite, E[T] = 1 / p provides a clear calculation method.The final answer is: boxed{1/p}

🔑:## Step 1: Define the forces acting on the massThe primary force acting on the mass is the force exerted by the spring, given by Hooke's Law as (F = -kx), where (k) is the spring constant and (x) is the displacement from the equilibrium position. Since the motion is considered in one dimension, we take the direction of positive (x) as the direction away from the equilibrium position. The force due to the spring is directed towards the equilibrium position, hence the negative sign.## Step 2: Apply Newton's second law of motionNewton's second law of motion states that the sum of all forces acting on an object equals its mass times its acceleration ((F = ma)). In this case, the only force acting on the mass is the spring force, so we have (F = -kx = ma).## Step 3: Express acceleration in terms of displacementAcceleration (a) is the second derivative of displacement (x) with respect to time (t), denoted as (frac{d^2x}{dt^2}). Therefore, we can rewrite Newton's second law for this system as (-kx = mfrac{d^2x}{dt^2}).## Step 4: Derive the differential equation of motionRearranging the equation from Step 3 to isolate the acceleration term on one side gives us (mfrac{d^2x}{dt^2} + kx = 0). This is a second-order linear homogeneous differential equation, which describes the motion of the mass-spring system.## Step 5: Simplify the differential equationDividing every term by (m) to simplify, we get (frac{d^2x}{dt^2} + frac{k}{m}x = 0). This equation can also be written as (frac{d^2x}{dt^2} + omega^2x = 0), where (omega^2 = frac{k}{m}) and (omega) is the angular frequency of the oscillation.The final answer is: boxed{frac{d^2x}{dt^2} + frac{k}{m}x = 0}