Fisica 3 Resolução Lista 4 Ótica

n_1 = 1\\n\\n\\theta\\text{ (t)}\\n\\n1) d_1 = d \\text{ y }\\n\\n2) \\sin \\theta_1 = \\sin \\theta_2\\n\\theta_j\\n\\hspace{2cm}\\n\\cos \\theta_2 \\sin \\theta_1\\n\\text{Se observa: }\\n\\sin \\theta_j\\sin \\theta_1\\n\\hspace{0.5cm}\\n\\text{(Encima de )}\\n\\cos \\theta_2\\sin \\theta_1\\n\\hspace{1.5cm}\\nd_1 = d_2\\sin \\theta_1\\sin \\theta_2\\n\\cos \\theta_1 = d\\sin \\theta_2\\n\\sin \\theta 2 = 0\\n\\u2026 d\\n\\n\\n\\n\\\\\\n\\n\\hspace{0.5cm}\\n\\sin \\theta_1 = 0\\n\\sin \\theta_1 = \\frac{c}{h}\\n\\sin \\theta_2 = \\frac{d}{d\\cos \\theta_1}\\n\\sin\\theta_2=\\sin \\alpha\\n\\num{3}\\n\\n\\sin \\theta_l=0\\n\\n\\sin \\theta_j\\hspace{0.5cm}\\n\\node[]{\\ensuremath{d_j =\\dfrac{d_i^2}{d}}} D + \\varphi = 90°\\n\\beta + \\varphi = 90°\\n\\varphi = 90° - \\theta\\n\\varphi = 90° - \\theta\\n\\beta - \\theta\\n\\n180° - \\varphi + \\alpha = 90° - \\theta + \\varepsilon\\n180° - 90° + \\alpha = \\varepsilon\\n\\n\\n\\n\\\\\\n\\n\\n\\n\\n\\\\\\n\\n\\n\\n\\n\\n\\n\\n\\\\\\n\\n\\n\\n\\\\\\n\\text{Uso:}\\n\\n3) v = 2m/s\\n\\theta = 30°\\n\\n\\\\n\\u2206\\alpha=\\tau_{n_1,n_2}\\n\\text{En terminos de }\\alpha\\text{, de }\\beta.\\n\\n\\n\\nb) E_c = \\sin^{-1}\\left(\\frac{n_2}{n_1}\\right)\\sin\\left(\\frac{4}{3}\\right)\\n=\\sin^{-1}(0.375)\\n\\theta_2 = 22.024° p = -i\nA uma dist. de 40 cm\na) Adotando o objeto como referência, a valor... seria como se o obj. estivesse parado em razão com o óbvio da colocada, v = √2v\nb1) Preferência em passos\n\\u2212\ns\nd\nd\na) θ = 40°\nn1 = 360° - 1\n\nn1 = 360°\n/n1 = 360° - 90\n\nn1 = 5/7\nd) θ = 100° f = k + r = 17.5\nma = -i/p\nhr - 25h\nhw = -25h\nmc = 0.75 = 1/p\n1/p = 1/p - 25p\n1 = 1 - 0.75\n1 = 1 - 0.1\np = 10.5 cm\na6 = 1/p = 17.5\n3) a)\nnh = p\n1/p = 1\nnh = nt - nh\n1/c\n0 n2 = nh - 1/r\n0\nn2 = (nh - 1)2\n...\nn2 = 2\nnh = 2\n\nimpossível! p = -i\nA uma dist. de 40 cm\na) Adotando o objeto como referência, a valor... seria como se o obj. estivesse parado em razão com o óbvio da colocada, v = √2v\nb1) Preferência em passos\n\\u2212\ns\nd\nd\na) θ = 40°\nn1 = 360° - 1\n\nn1 = 360°\n/n1 = 360° - 90\n\nn1 = 5/7\nd) θ = 100° f = 17.5\nm = -i/p\nh = 25 h\n\ni = -65p\nm = -2.5 = -1/p\n1/p = 1/f = 1 - 2.5p\n\n1/p = -0.4\n1/p = 125\n\np^2 = 10.5 cm\n\na) n1/n2 = n2-h2/n2 - h1\n\ncomo as raios se paralelos\n\np = \\infty\n\ni = 2r\n\nb) n1/n1 = 1/r\nn2 = n2 - 1\nn2 - n2/r = 0\n\nn2 = n1 - 2\nz - 2n2 - n2\nn1 = 2\n\nimpossível! n1 = 1/r = 1 - 1/s\n\n(3)\n1 - 1/5 = 1/(1+0)\n1 = 1/f\n\ni = f\n5/6 f = 5\n\nw1 = largura faixa incident\nw2 = (l1/l2)W1 = largura faixa cong\n\nf*B = W1/F1\n \u2192 tg B = W2/F2\n\nW1/W2 = W2/F1\n\ntg B = W2/F2 = W2 - F2/W2 = F1\n\nigual lencento imagem?\n\np = 20 cm\nf = 75 cm\n1/f = 1/p - 1/20\n\ni = 20 cm\nthen 1 = 20 cm\n\nm = -i/p\nh = 20/50 = 0.4cm\n\n1/(3X10) = 0.2\n\np = 30 cm\nf = 50 cm\n1/20 =?\n\ni = -1/p\nm1 = m1*n(\(k-1)/p-1\n 1) \n\na) 1 + 1 + 1 = 4\n f i\n f i\n\ni = 0.5 + 1 \n m = 1 \n\ndistancia de rangos formados está a 3cm do foco\n\nm_i = 1\n p = 1 \n i = 0.6m\nb)\nw real i > 0\n\nc) m_o - i / p = - 0.6 / 3 < 0\n\n\nd) m_r = m_1 m_2 = 1 (0.6) / 3\n\n\n p 56